Leibnitz's rule, Wall's Formula Questions in English

(B) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{\int_0^x {\cos {t^2}dt} }}{x}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L'\text{Hospital's rule}$.
Using the $\text{Leibniz integral rule}$,the derivative of the numerator $\frac{d}{dx} \int_0^x \cos(t^2) dt = \cos(x^2)$.
The derivative of the denominator $\frac{d}{dx} (x) = 1$.
Therefore,$L = \mathop {\lim }\limits_{x \to 0} \frac{\cos(x^2)}{1} = \cos(0) = 1$.

(C) Given $F(x) = \int_0^{x^2} f(t) dt = x^2(1 + x)$.
Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$:
$\frac{d}{dx} \left( \int_0^{x^2} f(t) dt \right) = \frac{d}{dx} (x^2 + x^3)$.
Using the chain rule,$f(x^2) \cdot \frac{d}{dx}(x^2) = 2x + 3x^2$.
$f(x^2) \cdot 2x = 2x + 3x^2$.
For $x > 0$,we can divide by $2x$:
$f(x^2) = \frac{2x + 3x^2}{2x} = 1 + \frac{3}{2}x$.
To find $f(4)$,we set $x^2 = 4$,which implies $x = 2$ (since $x > 0$):
$f(4) = 1 + \frac{3}{2}(2) = 1 + 3 = 4$.

(B) According to the Leibniz Integral Rule,if $f(x) = \int_{g(x)}^{h(x)} F(t, x) \, dt$,then $\frac{d}{dx} f(x) = F(h(x), x) \cdot h'(x) - F(g(x), x) \cdot g'(x)$.
Given $f(x) = \int_a^x t^3 e^t \, dt$.
Here,the integrand is $F(t) = t^3 e^t$,the upper limit is $h(x) = x$,and the lower limit is $g(x) = a$.
Applying the rule:
$\frac{d}{dx} f(x) = (x^3 e^x) \cdot \frac{d}{dx}(x) - (a^3 e^a) \cdot \frac{d}{dx}(a)$.
Since $a$ is a constant,$\frac{d}{dx}(a) = 0$.
Therefore,$\frac{d}{dx} f(x) = x^3 e^x \cdot 1 - 0 = x^3 e^x$.

(A) Given the equation: $\int_{\sin x}^1 {t^2}f(t) dt = 1 - \sin x$.
Applying the Leibniz integral rule to differentiate both sides with respect to $x$:
$\frac{d}{dx} \left( \int_{\sin x}^1 {t^2}f(t) dt \right) = \frac{d}{dx} (1 - \sin x)$.
Using the Fundamental Theorem of Calculus,the derivative of $\int_{g(x)}^a h(t) dt$ is $-h(g(x)) \cdot g'(x)$.
So,$-(\sin x)^2 f(\sin x) \cdot \cos x = -\cos x$.
Since $x \in (0, \frac{\pi}{2})$,$\cos x \neq 0$,we can divide both sides by $-\cos x$:
$(\sin x)^2 f(\sin x) = 1$.
Thus,$f(\sin x) = \frac{1}{\sin^2 x}$.
Replacing $\sin x$ with $t$,we get $f(t) = \frac{1}{t^2}$.
Therefore,$f\left( \frac{1}{\sqrt{3}} \right) = \frac{1}{(1/\sqrt{3})^2} = \frac{1}{1/3} = 3$.

(B) We use the Wallis formula for the definite integral $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.
Here,$m = 2$ and $n = 3$.
Substituting these values,we get:
$\int_0^{\pi/2} \sin^2 x \cos^3 x \, dx = \frac{\Gamma(\frac{2+1}{2}) \Gamma(\frac{3+1}{2})}{2 \Gamma(\frac{2+3+2}{2})} = \frac{\Gamma(\frac{3}{2}) \Gamma(2)}{2 \Gamma(\frac{7}{2})}$.
We know that $\Gamma(\frac{3}{2}) = \frac{1}{2} \sqrt{\pi}$,$\Gamma(2) = 1! = 1$,and $\Gamma(\frac{7}{2}) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{15}{8} \sqrt{\pi}$.
Thus,the integral is $\frac{(\frac{1}{2} \sqrt{\pi}) \cdot 1}{2 \cdot (\frac{15}{8} \sqrt{\pi})} = \frac{\frac{1}{2} \sqrt{\pi}}{\frac{15}{4} \sqrt{\pi}} = \frac{1}{2} \cdot \frac{4}{15} = \frac{2}{15}$.

(A) We need to evaluate $I = \int_0^\pi {\left| {\sin^4 x} \right|\,dx}$.
Since $\sin^4 x \ge 0$ for all $x \in [0, \pi]$,we can remove the absolute value sign: $I = \int_0^\pi \sin^4 x \,dx$.
Using the property $\int_0^{2a} f(x) \,dx = 2 \int_0^a f(x) \,dx$ if $f(2a-x) = f(x)$,we have:
$I = 2 \int_0^{\pi/2} \sin^4 x \,dx$.
Using Wallis' Formula,$\int_0^{\pi/2} \sin^n x \,dx = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ for even $n$:
$I = 2 \cdot \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \right) = 2 \cdot \frac{3\pi}{16} = \frac{3\pi}{8}$.

(B) Using Wallis's Formula for the integral of $\sin^{n} x$ from $0$ to $\pi/2$ where $n = 2m$ is even:
$\int_0^{\pi /2} \sin^{2m} x \, dx = \frac{2m-1}{2m} \cdot \frac{2m-3}{2m-2} \cdot \dots \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}$
To simplify this,multiply the numerator and denominator by the product of even numbers $2m \cdot (2m-2) \cdot \dots \cdot 2$:
Numerator: $(2m-1)(2m-3)\dots(1) \cdot [2m(2m-2)\dots(2)] = (2m)!$
Denominator: $[2m(2m-2)\dots(2)]^2 = [2^m \cdot m(m-1)\dots(1)]^2 = (2^m \cdot m!)^2$
Thus,the integral is $\frac{(2m)!}{(2^m \cdot m!)^2} \cdot \frac{\pi}{2}$.

(A) To evaluate the integral $I = \int_0^{\pi /2} \sin^5 x \, dx$,we use Wallis' Formula.
Wallis' Formula states that for $n > 1$:
$\int_0^{\pi /2} \sin^n x \, dx = \frac{(n-1)(n-3)\dots(1)}{n(n-2)\dots(2)} \times \frac{\pi}{2}$ (if $n$ is even)
$\int_0^{\pi /2} \sin^n x \, dx = \frac{(n-1)(n-3)\dots(2)}{n(n-2)\dots(3)}$ (if $n$ is odd)
Here,$n = 5$,which is an odd number.
Applying the formula:
$I = \frac{(5-1)(5-3)}{5(5-2)(5-4)} = \frac{4 \times 2}{5 \times 3 \times 1} = \frac{8}{15}$.
Thus,the correct option is $A$.

(C) Using the Leibniz integral rule,if $\varphi(x) = \int_{g(x)}^{h(x)} f(t) \, dt$,then $\varphi'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.
Here,$f(t) = \sin(t^2)$,$h(x) = \sqrt{x}$,and $g(x) = \frac{1}{x}$.
Calculating the derivatives: $h'(x) = \frac{1}{2\sqrt{x}}$ and $g'(x) = -\frac{1}{x^2}$.
Substituting these into the formula:
$\varphi'(x) = \sin((\sqrt{x})^2) \cdot \frac{1}{2\sqrt{x}} - \sin((\frac{1}{x})^2) \cdot (-\frac{1}{x^2})$
$\varphi'(x) = \sin(x) \cdot \frac{1}{2\sqrt{x}} + \frac{1}{x^2} \sin(\frac{1}{x^2})$.
Now,evaluate at $x = 1$:
$\varphi'(1) = \sin(1) \cdot \frac{1}{2\sqrt{1}} + \frac{1}{1^2} \sin(\frac{1}{1^2})$
$\varphi'(1) = \frac{1}{2} \sin(1) + \sin(1) = \frac{3}{2} \sin(1)$.

(B) Let $I = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx$.
Split the integral into two parts: $I = \int_{-\pi/2}^{\pi/2} \sin^3 x \cos^2 x \, dx + \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx$.
The first integral $\int_{-\pi/2}^{\pi/2} \sin^3 x \cos^2 x \, dx$ is an odd function,so its value is $0$.
The second integral $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx$ is an even function,so it equals $2 \int_0^{\pi/2} \sin^2 x \cos^3 x \, dx$.
Using Wallis' formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!}$ for even $m, n$:
$2 \int_0^{\pi/2} \sin^2 x \cos^3 x \, dx = 2 \times \frac{(2-1)!!(3-1)!!}{(2+3)!!} = 2 \times \frac{1 \times 2}{5 \times 3 \times 1} = 2 \times \frac{2}{15} = \frac{4}{15}$.

(D) We use the Leibniz Integral Rule: $\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) \, dt = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.
Given $F(x) = \int_{x^2}^{x^3} \frac{1}{\log t} \, dt$,we have $f(t) = \frac{1}{\log t}$,$h(x) = x^3$,and $g(x) = x^2$.
Then $h'(x) = 3x^2$ and $g'(x) = 2x$.
Applying the rule:
$F'(x) = \frac{1}{\log(x^3)} \cdot (3x^2) - \frac{1}{\log(x^2)} \cdot (2x)$
Since $\log(x^n) = n \log x$,we get:
$F'(x) = \frac{3x^2}{3 \log x} - \frac{2x}{2 \log x}$
$F'(x) = \frac{x^2}{\log x} - \frac{x}{\log x} = \frac{x^2 - x}{\log x} = \frac{x(x - 1)}{\log x} = x(x - 1)(\log x)^{-1}$.

(B) To find $f'(x)$ for the integral $f(x) = \int_{g(x)}^{h(x)} \phi(t) \, dt$,we use the Leibniz integral rule:
$f'(x) = \phi(h(x)) \cdot h'(x) - \phi(g(x)) \cdot g'(x)$
Here,$\phi(t) = \sin \sqrt{t}$,$h(x) = x^4$,and $g(x) = x^2$.
Calculating the derivatives:
$h'(x) = \frac{d}{dx}(x^4) = 4x^3$
$g'(x) = \frac{d}{dx}(x^2) = 2x$
Substituting these into the formula:
$f'(x) = \sin(\sqrt{x^4}) \cdot (4x^3) - \sin(\sqrt{x^2}) \cdot (2x)$
Since $\sqrt{x^4} = x^2$ and $\sqrt{x^2} = |x|$,assuming $x > 0$ for simplicity in standard calculus problems,we get:
$f'(x) = 4x^3 \sin(x^2) - 2x \sin(x)$
Thus,the correct option is $B$.

(C) Given $F(x) = \frac{1}{x^2} \int_4^x (4t^2 - 2F'(t)) \, dt$.
Multiplying by $x^2$,we get $x^2 F(x) = \int_4^x (4t^2 - 2F'(t)) \, dt$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$2x F(x) + x^2 F'(x) = 4x^2 - 2F'(x)$.
At $x = 4$:
$2(4) F(4) + (4)^2 F'(4) = 4(4)^2 - 2F'(4)$.
From the original equation,$F(4) = \frac{1}{4^2} \int_4^4 (4t^2 - 2F'(t)) \, dt = 0$.
Substituting $F(4) = 0$ into the differentiated equation:
$8(0) + 16 F'(4) = 64 - 2F'(4)$.
$16 F'(4) + 2 F'(4) = 64$.
$18 F'(4) = 64$.
$F'(4) = \frac{64}{18} = \frac{32}{9}$.

(A) Let $I = \int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$.
Substitute $t = \frac{x}{2}$,then $dx = 2 \, dt$.
When $x = 0$,$t = 0$. When $x = \pi$,$t = \frac{\pi}{2}$.
So,$I = \int_0^{\pi/2} \sin^5(t) \cdot 2 \, dt = 2 \int_0^{\pi/2} \sin^5(t) \, dt$.
Using Wallis' Formula,$\int_0^{\pi/2} \sin^n(x) \, dx = \frac{(n-1)(n-3)\dots(1)}{n(n-2)\dots(2)} \cdot \frac{\pi}{2}$ (if $n$ is even) or $\frac{(n-1)(n-3)\dots(2)}{n(n-2)\dots(1)}$ (if $n$ is odd).
For $n = 5$,$\int_0^{\pi/2} \sin^5(t) \, dt = \frac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{8}{15}$.
Therefore,$I = 2 \cdot \frac{8}{15} = \frac{16}{15}$.

(B) Using the Wallis formula for the integral $I = \int_0^{\pi /2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.
Here,$m = 4$ and $n = 6$.
$I = \frac{\Gamma(\frac{4+1}{2}) \Gamma(\frac{6+1}{2})}{2 \Gamma(\frac{4+6+2}{2})} = \frac{\Gamma(5/2) \Gamma(7/2)}{2 \Gamma(6)}$.
Using $\Gamma(n+1) = n\Gamma(n)$ and $\Gamma(1/2) = \sqrt{\pi}$:
$\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$.
$\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8}$.
$\Gamma(6) = 5! = 120$.
$I = \frac{(\frac{3\sqrt{\pi}}{4}) (\frac{15\sqrt{\pi}}{8})}{2 \cdot 120} = \frac{45\pi}{32 \cdot 240} = \frac{45\pi}{7680} = \frac{3\pi}{512}$.

(A) Given $F(x) = \int_{x^2}^{x^3} \log t \, dt$.
Applying the Leibniz integral rule,which states that $\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) \, dt = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.
Here,$f(t) = \log t$,$h(x) = x^3$,and $g(x) = x^2$.
Thus,$F'(x) = \log(x^3) \cdot \frac{d}{dx}(x^3) - \log(x^2) \cdot \frac{d}{dx}(x^2)$.
$F'(x) = (3 \log x) \cdot (3x^2) - (2 \log x) \cdot (2x)$.
$F'(x) = 9x^2 \log x - 4x \log x$.
$F'(x) = (9x^2 - 4x) \log x$.

(C) Let $I = \int_{-\pi/2}^{\pi/2} \sin^4 x \cos^6 x \, dx$.
Since $f(x) = \sin^4 x \cos^6 x$ is an even function,$f(-x) = f(x)$,we have:
$I = 2 \int_0^{\pi/2} \sin^4 x \cos^6 x \, dx$.
Using the Wallis formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{[(m-1)(m-3)...(1)][(n-1)(n-3)...(1)]}{(m+n)(m+n-2)...(2)} \times \frac{\pi}{2}$ (if both $m, n$ are even):
Here $m=4$ and $n=6$.
$I = 2 \times \left[ \frac{(3 \times 1) \times (5 \times 3 \times 1)}{(10 \times 8 \times 6 \times 4 \times 2)} \times \frac{\pi}{2} \right]$.
$I = 2 \times \left[ \frac{3 \times 15}{3840} \times \frac{\pi}{2} \right] = \frac{45\pi}{3840} = \frac{3\pi}{256}$.

(A) Given $f(x) = \int_{1}^{x} \sqrt{2 - t^2} dt$.
Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx} \int_{1}^{x} \sqrt{2 - t^2} dt = \sqrt{2 - x^2}$.
The given equation is $x^2 - f'(x) = 0$,which implies $x^2 = f'(x)$.
Substituting $f'(x)$,we get $x^2 = \sqrt{2 - x^2}$.
Squaring both sides,we obtain $x^4 = 2 - x^2$,which simplifies to $x^4 + x^2 - 2 = 0$.
Let $u = x^2$. Then the equation becomes $u^2 + u - 2 = 0$.
Factoring the quadratic,we get $(u + 2)(u - 1) = 0$.
Since $u = x^2$,we have $x^2 = -2$ (which gives no real roots) or $x^2 = 1$.
Thus,$x = \pm 1$.

(C) Let $x = a(1 - \cos \theta)$. Then $dx = a \sin \theta d\theta$.
When $x=0, \theta=0$ and when $x=a, \theta=\pi/2$.
The expression $2ax - x^2 = 2a^2(1 - \cos \theta) - a^2(1 - \cos \theta)^2 = a^2(1 - \cos^2 \theta) = a^2 \sin^2 \theta$.
Substituting these into the integral:
$\int_0^a x(2ax - x^2)^{3/2} dx = \int_0^{\pi/2} a(1 - \cos \theta) (a^2 \sin^2 \theta)^{3/2} a \sin \theta d\theta$
$= a^5 \int_0^{\pi/2} (1 - \cos \theta) \sin^4 \theta d\theta$
$= a^5 [\int_0^{\pi/2} \sin^4 \theta d\theta - \int_0^{\pi/2} \sin^4 \theta \cos \theta d\theta]$
Using Wallis' formula for the first part: $\int_0^{\pi/2} \sin^4 \theta d\theta = \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2} = \frac{3\pi}{16}$.
For the second part,let $u = \sin \theta$,then $du = \cos \theta d\theta$: $\int_0^1 u^4 du = [\frac{u^5}{5}]_0^1 = \frac{1}{5}$.
Thus,the result is $a^5 [\frac{3\pi}{16} - \frac{1}{5}]$.

(B) Given that $f(x) = \int_0^x {t\sin t\,dt}$.
According to the Leibniz integral rule,if $f(x) = \int_{g(x)}^{h(x)} F(t, x) \, dt$,then $f'(x) = F(h(x), x) \cdot h'(x) - F(g(x), x) \cdot g'(x) + \int_{g(x)}^{h(x)} \frac{\partial}{\partial x} F(t, x) \, dt$.
In this case,$F(t, x) = t\sin t$,$h(x) = x$,and $g(x) = 0$.
Since the integrand $t\sin t$ does not depend on $x$,the partial derivative $\frac{\partial}{\partial x} (t\sin t) = 0$.
Therefore,$f'(x) = (x\sin x) \cdot \frac{d}{dx}(x) - (0\sin 0) \cdot \frac{d}{dx}(0) + 0$.
$f'(x) = x\sin x \cdot 1 - 0 = x\sin x$.

(A) Given the equation: $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt}$
Rewrite the integral $\int_x^1 {t\,f(t)\,dt}$ as $-\int_1^x {t\,f(t)\,dt}$:
$\int_0^x {f(t)\,dt} = x - \int_1^x {t\,f(t)\,dt}$
Differentiating both sides with respect to $x$ using the Leibniz integral rule:
$\frac{d}{dx} \left( \int_0^x {f(t)\,dt} \right) = \frac{d}{dx} (x) - \frac{d}{dx} \left( \int_1^x {t\,f(t)\,dt} \right)$
Applying the Fundamental Theorem of Calculus:
$f(x) = 1 - xf(x)$
Rearranging the terms to solve for $f(x)$:
$f(x) + xf(x) = 1$
$f(x)(1 + x) = 1$
$f(x) = \frac{1}{1 + x}$
Now,substitute $x = 1$ to find $f(1)$:
$f(1) = \frac{1}{1 + 1} = \frac{1}{2}$.

(A) Given the equation $\int_0^{t^2} xf(x)dx = \frac{2}{5}t^5$.
Applying the Leibniz rule for differentiation under the integral sign with respect to $t$:
$\frac{d}{dt} \left( \int_0^{t^2} xf(x)dx \right) = \frac{d}{dt} \left( \frac{2}{5}t^5 \right)$.
Using the chain rule,we get:
$(t^2)f(t^2) \cdot \frac{d}{dt}(t^2) = \frac{2}{5} \cdot 5t^4$.
$(t^2)f(t^2) \cdot (2t) = 2t^4$.
$2t^3 f(t^2) = 2t^4$.
For $t > 0$,we can divide by $2t^3$:
$f(t^2) = t$.
We need to find $f\left( \frac{4}{25} \right)$.
Let $t^2 = \frac{4}{25}$,which implies $t = \frac{2}{5}$ (since $t > 0$).
Substituting $t = \frac{2}{5}$ into $f(t^2) = t$,we get:
$f\left( \frac{4}{25} \right) = \frac{2}{5}$.

(B) Given $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$.
Using Leibniz's rule for differentiation under the integral sign:
$f'(x) = e^{-(x^2+1)^2} \cdot \frac{d}{dx}(x^2+1) - e^{-(x^2)^2} \cdot \frac{d}{dx}(x^2)$
$f'(x) = e^{-(x^2+1)^2} \cdot (2x) - e^{-x^4} \cdot (2x)$
$f'(x) = 2x \left( e^{-(x^2+1)^2} - e^{-x^4} \right)$
Since $(x^2+1)^2 > x^4$ for all $x$,it follows that $-(x^2+1)^2 < -x^4$.
Therefore,$e^{-(x^2+1)^2} < e^{-x^4}$,which implies $\left( e^{-(x^2+1)^2} - e^{-x^4} \right) < 0$.
For $f(x)$ to be an increasing function,we require $f'(x) \ge 0$.
Since the term in the parenthesis is always negative,$f'(x) \ge 0$ only when $2x \le 0$,which means $x \le 0$.
Thus,$f(x)$ is increasing in the interval $(-\infty, 0)$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x}$.
Since this is a $\frac{0}{0}$ form,we apply $L'\text{H\^opital's rule}$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(\int_0^{x^2} \sec^2 t \, dt)}{\frac{d}{dx}(x \sin x)}$.
Using the $Leibniz$ rule for differentiation under the integral sign:
$L = \mathop {\lim }\limits_{x \to 0} \frac{\sec^2(x^2) \cdot 2x}{\sin x + x \cos x}$.
Divide numerator and denominator by $x$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{2 \sec^2(x^2)}{\frac{\sin x}{x} + \cos x}$.
Substituting $x = 0$:
$L = \frac{2 \cdot \sec^2(0)}{1 + \cos(0)} = \frac{2 \cdot 1}{1 + 1} = \frac{2}{2} = 1$.

(C) Given $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cos^2 x \end{array} \right|$.
Applying $R_1 \to R_1 - \sec x R_3$,the first row becomes:
$R_1 = (\sec x - \sec x, \cos x - \sec x \cos^2 x, \sec^2 x + \cot x \csc x - \sec x \cos^2 x)$
$R_1 = (0, 0, \sec^2 x + \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} - \cos x) = (0, 0, \sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)$.
Expanding along the first row,$f(x) = (\sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)(\cos^4 x - \cos^2 x) = (\sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)(-\cos^2 x \sin^2 x) = -\sin^2 x - \cos^5 x + \cos^3 x \sin^2 x$.
Actually,simplifying the determinant directly: $f(x) = -\sin^2 x - \cos^5 x$.
Now,$\int_0^{\pi/2} f(x) dx = -\int_0^{\pi/2} (\sin^2 x + \cos^5 x) dx$.
Using Wallis formula: $\int_0^{\pi/2} \sin^2 x dx = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$.
$\int_0^{\pi/2} \cos^5 x dx = \frac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{8}{15}$.
Therefore,$\int_0^{\pi/2} f(x) dx = -(\frac{\pi}{4} + \frac{8}{15}) = -\frac{\pi}{4} - \frac{8}{15}$.

(D) Let $F(x) = \int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$.
Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:
$F'(x) = \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} \cdot \frac{d}{dx}(x^2)$
$F'(x) = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot 2x$
For extremum points,we set $F'(x) = 0$:
$\frac{(x^2 - 1)(x^2 - 4)}{2 + e^{x^2}} \cdot 2x = 0$
This implies $x = 0$ or $x^2 = 1$ or $x^2 = 4$.
Solving these,we get $x = 0$,$x = \pm 1$,and $x = \pm 2$.
Since all these values are included in the options provided,the correct answer is $D$.

(B) Given the equation: $\int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du + \int_0^y \cos t \,dt = 0.$
Applying the Leibniz rule for differentiation under the integral sign with respect to $x$:
$\frac{d}{dx} \left( \int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du \right) + \frac{d}{dx} \left( \int_0^y \cos t \,dt \right) = 0.$
Using the Fundamental Theorem of Calculus:
$\sqrt{3 - 2\sin^2 x} + \cos y \cdot \frac{dy}{dx} = 0.$
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\cos y \cdot \frac{dy}{dx} = -\sqrt{3 - 2\sin^2 x}.$
Therefore,$\frac{dy}{dx} = -\frac{\sqrt{3 - 2\sin^2 x}}{\cos y}.$
Thus,the correct option is $B$.

(C) Let $L = \mathop {Lim}\limits_{k \to 0} \frac{\int\limits_0^k (1 + \sin 2x)^{\frac{1}{x}} dx}{k}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'Hopital's rule by differentiating the numerator and the denominator with respect to $k$.
Using the Fundamental Theorem of Calculus (Leibnitz's rule),the derivative of the numerator is $(1 + \sin 2k)^{\frac{1}{k}}$ and the derivative of the denominator is $1$.
Thus,$L = \mathop {Lim}\limits_{k \to 0} (1 + \sin 2k)^{\frac{1}{k}}$.
This is of the form $1^{\infty}$,which can be evaluated as $e^{\mathop {Lim}\limits_{k \to 0} \frac{1}{k} \ln(1 + \sin 2k)}$.
Using the standard limit $\mathop {Lim}\limits_{u \to 0} \frac{\ln(1+u)}{u} = 1$,we have $\mathop {Lim}\limits_{k \to 0} \frac{\ln(1 + \sin 2k)}{k} = \mathop {Lim}\limits_{k \to 0} \left( \frac{\ln(1 + \sin 2k)}{\sin 2k} \cdot \frac{\sin 2k}{2k} \cdot 2 \right) = 1 \cdot 1 \cdot 2 = 2$.
Therefore,$L = e^2$.

(C) Let $I = \int\limits_0^{\frac{\pi}{4}} (\cos 2x)^{3/2} \cos x \,dx$.
Using the identity $\cos 2x = 1 - 2\sin^2 x$,we have:
$I = \int\limits_0^{\frac{\pi}{4}} (1 - 2\sin^2 x)^{3/2} \cos x \,dx$.
Let $\sqrt{2} \sin x = \sin \theta$. Then $\sqrt{2} \cos x \,dx = \cos \theta \,d\theta$,which implies $\cos x \,dx = \frac{1}{\sqrt{2}} \cos \theta \,d\theta$.
When $x = 0$,$\sin \theta = 0 \implies \theta = 0$. When $x = \frac{\pi}{4}$,$\sin \theta = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \implies \theta = \frac{\pi}{2}$.
Substituting these into the integral:
$I = \int\limits_0^{\frac{\pi}{2}} (1 - \sin^2 \theta)^{3/2} \cdot \frac{1}{\sqrt{2}} \cos \theta \,d\theta = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} (\cos^2 \theta)^{3/2} \cos \theta \,d\theta$.
$I = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} \cos^3 \theta \cdot \cos \theta \,d\theta = \frac{1}{\sqrt{2}} \int\limits_0^{\frac{\pi}{2}} \cos^4 \theta \,d\theta$.
Using Wallis' formula $\int\limits_0^{\frac{\pi}{2}} \cos^n \theta \,d\theta = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ for even $n$:
$I = \frac{1}{\sqrt{2}} \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \right) = \frac{1}{\sqrt{2}} \cdot \frac{3\pi}{16} = \frac{3\pi}{16\sqrt{2}}$.

(D) Given $f(x) = 1 + x + \int\limits_1^x (\ln^2 t + 2 \ln t) \, dt$.
Applying Leibniz's rule to differentiate with respect to $x$:
$f'(x) = 0 + 1 + (\ln^2 x + 2 \ln x) = 0$.
This simplifies to $(\ln x + 1)^2 = 0$,which gives $\ln x = -1$,so $x = e^{-1}$.
Now,evaluate $f(e^{-1}) = 1 + e^{-1} + \int\limits_1^{e^{-1}} (\ln^2 t + 2 \ln t) \, dt$.
Note that $\frac{d}{dt}(t \ln^2 t) = \ln^2 t + t(2 \ln t \cdot \frac{1}{t}) = \ln^2 t + 2 \ln t$.
Thus,$\int (\ln^2 t + 2 \ln t) \, dt = t \ln^2 t + C$.
Evaluating the definite integral:
$f(e^{-1}) = 1 + e^{-1} + [t \ln^2 t]_1^{e^{-1}} = 1 + e^{-1} + (e^{-1} \cdot (-1)^2 - 1 \cdot 0^2) = 1 + e^{-1} + e^{-1} = 1 + 2e^{-1}$.

(A) Given the equation: $x \cdot \sin(\pi x) = \int_{0}^{x^2} f(t) \, dt$.
Applying the Leibniz rule for differentiation under the integral sign with respect to $x$:
$\frac{d}{dx} [x \cdot \sin(\pi x)] = \frac{d}{dx} \int_{0}^{x^2} f(t) \, dt$.
Using the product rule on the left side and the Fundamental Theorem of Calculus on the right side:
$1 \cdot \sin(\pi x) + x \cdot \cos(\pi x) \cdot \pi = f(x^2) \cdot \frac{d}{dx}(x^2)$.
$\sin(\pi x) + \pi x \cos(\pi x) = f(x^2) \cdot 2x$.
To find $f(4)$,we set $x^2 = 4$,which implies $x = 2$ or $x = -2$.
Case $1$: Let $x = 2$.
$\sin(2\pi) + \pi(2) \cos(2\pi) = f(4) \cdot 2(2)$.
$0 + 2\pi(1) = 4 \cdot f(4)$.
$2\pi = 4 \cdot f(4) \implies f(4) = \frac{\pi}{2}$.
Case $2$: Let $x = -2$.
$\sin(-2\pi) + \pi(-2) \cos(-2\pi) = f(4) \cdot 2(-2)$.
$0 - 2\pi(1) = -4 \cdot f(4)$.
$-2\pi = -4 \cdot f(4) \implies f(4) = \frac{\pi}{2}$.
Thus,the value of $f(4)$ is $\frac{\pi}{2}$.

(B) Let the given limit be $L = \mathop {Limit}\limits_{x \to {x_1}} \frac{x \int_{x_1}^x f(t) dt}{x - x_1}$.
Since the form is $\frac{0}{0}$ as $x \to x_1$,we apply $L$'$H$ôpital's rule.
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx} [x \int_{x_1}^x f(t) dt] = 1 \cdot \int_{x_1}^x f(t) dt + x \cdot f(x)$ (using Leibniz's rule).
Denominator derivative: $\frac{d}{dx} (x - x_1) = 1$.
Thus,$L = \mathop {Limit}\limits_{x \to {x_1}} [\int_{x_1}^x f(t) dt + x f(x)]$.
As $x \to x_1$,$\int_{x_1}^{x_1} f(t) dt = 0$.
Therefore,$L = 0 + x_1 f(x_1) = x_1 f(x_1)$.
The correct option is $B$.

(B) Using the Leibniz integral rule,we differentiate $S(x)$ with respect to $x$:
$S'(x) = \ln(x^3) \cdot \frac{d}{dx}(x^3) - \ln(x^2) \cdot \frac{d}{dx}(x^2)$
$S'(x) = (3 \ln x) \cdot (3x^2) - (2 \ln x) \cdot (2x)$
$S'(x) = 9x^2 \ln x - 4x \ln x$
$S'(x) = x \ln x (9x - 4)$
Now,$H(x) = \frac{S'(x)}{x} = \frac{x \ln x (9x - 4)}{x} = \ln x (9x - 4)$
Since $\ln x$ and $(9x - 4)$ are both continuous and differentiable for all $x > 0$,their product $H(x) = \ln x (9x - 4)$ is also continuous and differentiable in its domain $(0, \infty)$.

(D) Let $f(x) = \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}}$.
Using Leibniz's rule,$\frac{d}{dx} f(x) = \frac{1}{\sqrt{1 - (\sin x)^2}} \cdot \frac{d}{dx}(\sin x) - \frac{1}{\sqrt{1 - (\cos x)^2}} \cdot \frac{d}{dx}(\cos x)$.
Since $\sqrt{1 - \sin^2 x} = |\cos x|$ and $\sqrt{1 - \cos^2 x} = |\sin x|$,the expression becomes $\frac{\cos x}{|\cos x|} - \frac{-\sin x}{|\sin x|} = \text{sgn}(\cos x) + \text{sgn}(\sin x)$.
In the interval $[0, \pi]$,$\sin x$ is always positive,so $\text{sgn}(\sin x) = 1$.
Thus,the equation is $\text{sgn}(\cos x) + 1 = 2\sqrt{2}$.
Since $\text{sgn}(\cos x)$ can only be $1, -1,$ or $0$,the sum $\text{sgn}(\cos x) + 1$ can only be $2, 0,$ or $1$.
Since $2\sqrt{2} \approx 2.828$,there is no value of $x$ such that $\text{sgn}(\cos x) + 1 = 2\sqrt{2}$.
Therefore,the number of solutions is $0$.

(C) Let $F(x) = \int\limits_x^{x + 3} {t(5 - t)\,dt}$.
Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:
$F'(x) = (x + 3)(5 - (x + 3)) \cdot \frac{d}{dx}(x + 3) - x(5 - x) \cdot \frac{d}{dx}(x)$
$F'(x) = (x + 3)(2 - x) - x(5 - x)$
$F'(x) = (2x - x^2 + 6 - 3x) - (5x - x^2)$
$F'(x) = 6 - x - x^2 - 5x + x^2$
$F'(x) = 6 - 6x$.
To find the critical points,set $F'(x) = 0$:
$6 - 6x = 0 \Rightarrow x = 1$.
Now,find the second derivative to check for maxima:
$F''(x) = -6$.
Since $F''(1) = -6 < 0$,the function $F(x)$ has a local maximum at $x = 1$.

(A) Given function: $f(x) = \sin x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta$.
Using Leibniz's rule for differentiation:
$f'(x) = \cos x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta + \sin x \left[ \frac{\sin \sqrt{x^2}}{1 + \cos^2 \sqrt{x^2}} \cdot \frac{d}{dx}(x^2) \right]$.
$f'(x) = \cos x \int_{\pi^2/16}^{x^2} \frac{\sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta + \sin x \left[ \frac{\sin x}{1 + \cos^2 x} \cdot 2x \right]$.
Now,substituting $x = \frac{\pi}{2}$:
$f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) \int_{\pi^2/16}^{\pi^2/4} \frac{\sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta + \sin(\frac{\pi}{2}) \left[ \frac{\sin(\frac{\pi}{2})}{1 + \cos^2(\frac{\pi}{2})} \cdot 2(\frac{\pi}{2}) \right]$.
Since $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$:
$f'(\frac{\pi}{2}) = 0 \cdot (\text{integral value}) + 1 \cdot \left[ \frac{1}{1 + 0} \cdot \pi \right] = \pi$.

(D) Let $I(x) = \int_0^{\pi/2} \frac{\ln(1 + x \sin^2 \theta)}{\sin^2 \theta} d\theta$.
Using Leibniz rule for differentiation under the integral sign:
$f'(x) = \int_0^{\pi/2} \frac{\partial}{\partial x} \left( \frac{\ln(1 + x \sin^2 \theta)}{\sin^2 \theta} \right) d\theta = \int_0^{\pi/2} \frac{1}{1 + x \sin^2 \theta} d\theta$.
Multiply numerator and denominator by $\sec^2 \theta$:
$f'(x) = \int_0^{\pi/2} \frac{\sec^2 \theta}{\sec^2 \theta + x \tan^2 \theta} d\theta = \int_0^{\pi/2} \frac{\sec^2 \theta}{1 + (1+x) \tan^2 \theta} d\theta$.
Let $u = \sqrt{1+x} \tan \theta$,then $du = \sqrt{1+x} \sec^2 \theta d\theta$.
$f'(x) = \frac{1}{\sqrt{1+x}} \int_0^{\infty} \frac{du}{1+u^2} = \frac{1}{\sqrt{1+x}} [\tan^{-1} u]_0^{\infty} = \frac{\pi}{2\sqrt{1+x}}$.
Integrating $f'(x)$ with respect to $x$:
$f(x) = \int \frac{\pi}{2\sqrt{1+x}} dx = \pi \sqrt{1+x} + C$.
Since $f(0) = \int_0^{\pi/2} \frac{\ln(1)}{\sin^2 \theta} d\theta = 0$,we have $\pi \sqrt{1} + C = 0 \implies C = -\pi$.
Thus,$f(x) = \pi(\sqrt{x+1} - 1)$.
Both $(A)$ and $(B)$ are correct.

(C) Given $f(x) = \int_{x}^{x^2} (t - 1) dt$.
Using the Leibniz rule for differentiation,$f'(x) = (x^2 - 1) \cdot \frac{d}{dx}(x^2) - (x - 1) \cdot \frac{d}{dx}(x)$.
$f'(x) = (x^2 - 1)(2x) - (x - 1)(1) = 2x^3 - 2x - x + 1 = 2x^3 - 3x + 1$.
Factoring $f'(x)$,we get $f'(x) = (x - 1)(2x^2 + 2x - 1)$.
For $1 \le x \le 2$,$f'(x) > 0$,which means $f(x)$ is a strictly increasing function on the interval $[1, 2]$.
Since $f(x)$ is strictly increasing,the global maximum value occurs at the right endpoint,$x = 2$.
$f(2) = \int_{2}^{4} (t - 1) dt = \left[ \frac{t^2}{2} - t \right]_{2}^{4} = (8 - 4) - (2 - 2) = 4 - 0 = 4$.

(B) Using the Leibniz integral rule,$f'(x) = \frac{d}{dx} \int_{g(x)}^{h(x)} F(t) dt = F(h(x)) \cdot h'(x) - F(g(x)) \cdot g'(x)$.
Here,$F(t) = 5^{\sqrt{t}}$,$h(x) = x^4$,and $g(x) = 9x^2$.
$f'(x) = 5^{\sqrt{x^4}} \cdot (4x^3) - 5^{\sqrt{9x^2}} \cdot (18x) = 5^{x^2} \cdot 4x^3 - 5^{3|x|} \cdot 18x$.
For $x = 3$,$f'(3) = 5^{3^2} \cdot 4(3^3) - 5^{3(3)} \cdot 18(3) = 5^9 \cdot 108 - 5^9 \cdot 54 = 54 \cdot 5^9$.
The limit is $\lim_{h \to 0} \frac{f(3 + h) - f(3 - h)}{h} = f'(3) - (-f'(3)) = 2f'(3)$.
$2 \cdot (54 \cdot 5^9) = 108 \cdot 5^9$.

(B) Let $I = \int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$. Since $f(\theta) = \sin^2 \theta \sin^2 2\theta$ is an even function,$I = 2 \int_{0}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$.
Using $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin^2 2\theta = 4 \sin^2 \theta \cos^2 \theta$.
So,$I = 2 \int_{0}^{3\pi} \sin^2 \theta (4 \sin^2 \theta \cos^2 \theta) \, d\theta = 8 \int_{0}^{3\pi} \sin^4 \theta \cos^2 \theta \, d\theta$.
Since the period of $\sin^4 \theta \cos^2 \theta$ is $\pi$,$\int_{0}^{3\pi} f(\theta) \, d\theta = 3 \int_{0}^{\pi} f(\theta) \, d\theta$.
Thus,$I = 8 \times 3 \int_{0}^{\pi} \sin^4 \theta \cos^2 \theta \, d\theta = 24 \times 2 \int_{0}^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta = 48 \int_{0}^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta$.
Using Wallis's formula $\int_{0}^{\pi/2} \sin^m \theta \cos^n \theta \, d\theta = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \times \frac{\pi}{2}$ (for even $m, n$):
$I = 48 \times \frac{(3 \times 1) \times (1)}{(6 \times 4 \times 2)} \times \frac{\pi}{2} = 48 \times \frac{3}{48} \times \frac{\pi}{2} = \frac{3\pi}{2}$.

(B) Let $I = \int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$. Since the integrand is an even function,$I = 2 \int_{0}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$.
Using $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin^2 2\theta = 4 \sin^2 \theta \cos^2 \theta$.
So,$I = 2 \int_{0}^{3\pi} \sin^2 \theta (4 \sin^2 \theta \cos^2 \theta) \, d\theta = 8 \int_{0}^{3\pi} \sin^4 \theta \cos^2 \theta \, d\theta$.
Since $\sin^4 \theta \cos^2 \theta$ has a period of $\pi$,$\int_{0}^{3\pi} f(\theta) \, d\theta = 3 \int_{0}^{\pi} f(\theta) \, d\theta$.
Thus,$I = 8 \times 3 \int_{0}^{\pi} \sin^4 \theta \cos^2 \theta \, d\theta = 24 \int_{0}^{\pi} \sin^4 \theta \cos^2 \theta \, d\theta$.
Using the property $\int_{0}^{\pi} f(\sin \theta) \, d\theta = 2 \int_{0}^{\pi/2} f(\sin \theta) \, d\theta$,we get $I = 48 \int_{0}^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta$.
Using Wallis's formula $\int_{0}^{\pi/2} \sin^m \theta \cos^n \theta \, d\theta = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \times \frac{\pi}{2}$ (for even $m, n$):
$I = 48 \times \frac{(3 \times 1) \times (1)}{(6 \times 4 \times 2)} \times \frac{\pi}{2} = 48 \times \frac{3}{48} \times \frac{\pi}{2} = \frac{3\pi}{2}$.

(C) Given $H(x) = (x + 1) \int_{x^2}^{x^3} \sin(t^3) dt$.
Since $H(1) = (1 + 1) \int_{1}^{1} \sin(t^3) dt = 0$,the limit $\lim_{x \to 1} \frac{H(x)}{x - 1}$ is of the form $\frac{0}{0}$.
Applying $L'\text{Hôpital's rule}$,we have $\lim_{x \to 1} \frac{H'(x)}{1} = H'(1)$.
Using the product rule and Leibniz integral rule:
$H'(x) = \frac{d}{dx}(x + 1) \cdot \int_{x^2}^{x^3} \sin(t^3) dt + (x + 1) \cdot \frac{d}{dx} \left( \int_{x^2}^{x^3} \sin(t^3) dt \right)$.
$H'(x) = 1 \cdot \int_{x^2}^{x^3} \sin(t^3) dt + (x + 1) \left[ \sin((x^3)^3) \cdot 3x^2 - \sin((x^2)^3) \cdot 2x \right]$.
Evaluating at $x = 1$:
$H'(1) = \int_{1}^{1} \sin(t^3) dt + (1 + 1) \left[ 3(1)^2 \sin(1^9) - 2(1) \sin(1^6) \right]$.
$H'(1) = 0 + 2 [3 \sin(1) - 2 \sin(1)] = 2 \sin(1)$.

(B) Given $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$.
Using the Leibniz rule for differentiation,$f'(x) = (x^2 - 1)(x^2 - 4)(x^2 - 9) \cdot \frac{d}{dx}(x^2) = 2x(x^2 - 1)(x^2 - 4)(x^2 - 9)$.
$f'(x) = 2x(x - 1)(x + 1)(x - 2)(x + 2)(x - 3)(x + 3)$.
The critical points are $x = 0, \pm 1, \pm 2, \pm 3$,which are $7$ distinct points.
$f'(x)$ is an odd function,so $f''(x)$ is an even function and $f'''(x)$ is an odd function.
Since $f'(x)$ is a polynomial of degree $7$,$f'''(x)$ is a polynomial of degree $5$.
As $f'''(x)$ is an odd function,$x = 0$ is a root of $f'''(x) = 0$.
By Rolle's Theorem,between any two roots of $f'(x)$,there is at least one root of $f''(x)$. Since $f'(x)$ has $7$ roots,$f''(x)$ has $6$ roots.
Similarly,$f'''(x)$ has $5$ roots. Since $f'''(x)$ is an odd function and $x=0$ is a root,the remaining $4$ roots must be symmetric about the origin,i.e.,$2$ positive and $2$ negative roots.
Thus,$f'''(x) = 0$ has $2$ distinct positive solutions.

(C) To find the intersection with the $x$-axis,we set $y = 0$. The integral $\int_{x^2}^{x^3} \sqrt{5 - t^2} \, dt = 0$ implies $x^2 = x^3$,so $x = 1$ (since $x \neq 0$).
Using the Leibniz Rule for differentiation,the slope of the tangent is given by $y' = \frac{d}{dx} \int_{x^2}^{x^3} \sqrt{5 - t^2} \, dt = \sqrt{5 - (x^3)^2} \cdot (3x^2) - \sqrt{5 - (x^2)^2} \cdot (2x)$.
At $x = 1$,the slope $m$ is $y'(1) = \sqrt{5 - 1} \cdot 3(1)^2 - \sqrt{5 - 1} \cdot 2(1) = 2 \cdot 3 - 2 \cdot 2 = 6 - 4 = 2$.
The angle $\theta$ between the curve and the $x$-axis is given by $\tan \theta = |m| = 2$.
Thus,$\theta = \tan^{-1} 2 = \cot^{-1} \frac{1}{2}$.

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) dx$.
Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$,we have:
$I = \int_{0}^{\frac{\pi}{2}} \left[ \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) + \sin^4(-x) \left( 1 + \log \left( \frac{2 + \sin(-x)}{2 - \sin(-x)} \right) \right) \right] dx$.
Since $\sin(-x) = -\sin x$,we get:
$I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \left[ 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) + 1 + \log \left( \frac{2 - \sin x}{2 + \sin x} \right) \right] dx$.
$I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \left[ 2 + \log \left( \frac{2 + \sin x}{2 - \sin x} \cdot \frac{2 - \sin x}{2 + \sin x} \right) \right] dx$.
$I = \int_{0}^{\frac{\pi}{2}} \sin^4 x [ 2 + \log(1) ] dx = \int_{0}^{\frac{\pi}{2}} 2 \sin^4 x dx = 2 \int_{0}^{\frac{\pi}{2}} \sin^4 x dx$.
Using Wallis' formula,$\int_{0}^{\frac{\pi}{2}} \sin^n x dx = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ for even $n$:
$I = 2 \cdot \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \right) = 2 \cdot \frac{3\pi}{16} = \frac{3\pi}{8}$.

(D) Given that $f : (-1, 1) \to R$ is a continuous function and $\int\limits_0^{\sin x} {f(t)dt} = \frac{\sqrt{3}}{2}x$.
Applying the Leibniz rule for differentiation under the integral sign on both sides with respect to $x$:
$\frac{d}{dx} \left( \int\limits_0^{\sin x} {f(t)dt} \right) = \frac{d}{dx} \left( \frac{\sqrt{3}}{2}x \right)$
$f(\sin x) \cdot \frac{d}{dx}(\sin x) = \frac{\sqrt{3}}{2}$
$f(\sin x) \cdot \cos x = \frac{\sqrt{3}}{2}$
To find $f\left(\frac{\sqrt{3}}{2}\right)$,we set $\sin x = \frac{\sqrt{3}}{2}$.
This implies $x = \frac{\pi}{3}$ (since $x \in (-1, 1)$ is satisfied by the domain of $\sin x$ and the integral).
Substituting $x = \frac{\pi}{3}$ into the equation:
$f\left(\sin \frac{\pi}{3}\right) \cdot \cos \frac{\pi}{3} = \frac{\sqrt{3}}{2}$
$f\left(\frac{\sqrt{3}}{2}\right) \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$
Multiplying both sides by $2$,we get:
$f\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}$

(A) Given the equation: $\int_{-\pi}^{t} (f(x) + x) dx = \pi^2 - t^2$
Splitting the integral,we get: $\int_{-\pi}^{t} f(x) dx + \int_{-\pi}^{t} x dx = \pi^2 - t^2$
Evaluating the second integral: $\int_{-\pi}^{t} x dx = \left[ \frac{x^2}{2} \right]_{-\pi}^{t} = \frac{t^2}{2} - \frac{(-\pi)^2}{2} = \frac{t^2}{2} - \frac{\pi^2}{2}$
Substituting this back: $\int_{-\pi}^{t} f(x) dx + \frac{t^2}{2} - \frac{\pi^2}{2} = \pi^2 - t^2$
Rearranging the terms: $\int_{-\pi}^{t} f(x) dx = \pi^2 - t^2 - \frac{t^2}{2} + \frac{\pi^2}{2} = \frac{3}{2}\pi^2 - \frac{3}{2}t^2 = \frac{3}{2}(\pi^2 - t^2)$
Differentiating both sides with respect to $t$ using the Leibniz rule: $\frac{d}{dt} \left[ \int_{-\pi}^{t} f(x) dx \right] = \frac{d}{dt} \left[ \frac{3}{2}(\pi^2 - t^2) \right]$
By the Fundamental Theorem of Calculus: $f(t) = \frac{3}{2}(0 - 2t) = -3t$
Therefore,$f\left(-\frac{\pi}{3}\right) = -3 \left(-\frac{\pi}{3}\right) = \pi$

(D) Given the equation $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$.
Applying the Leibniz integral rule,we differentiate both sides with respect to $x$:
$\frac{d}{{dx}}\left[ {\int\limits_e^x {t\,f(t)\,dt} } \right] = \frac{d}{{dx}}\left[ {\sin x - x\cos x - \frac{{{x^2}}}{2}} \right]$
Using the product rule on the right side:
$x\,f(x) = \cos x - (\cos x - x\sin x) - x$
$x\,f(x) = \cos x - \cos x + x\sin x - x$
$x\,f(x) = x\sin x - x$
Dividing by $x$ (since $x \neq 0$):
$f(x) = \sin x - 1$
Now,substitute $x = \frac{\pi}{6}$:
$f(\frac{\pi}{6}) = \sin(\frac{\pi}{6}) - 1$
$f(\frac{\pi}{6}) = \frac{1}{2} - 1 = -\frac{1}{2}$.

(B) Let $I = \int \limits_{0}^{\pi} |\cos x|^3 dx$.
Since the function $f(x) = |\cos x|^3$ is symmetric about $x = \frac{\pi}{2}$ in the interval $[0, \pi]$,we can write:
$I = 2 \int \limits_{0}^{\frac{\pi}{2}} \cos^3 x dx$.
Using Wallis' formula,$\int \limits_{0}^{\frac{\pi}{2}} \cos^n x dx = \frac{(n-1)!!}{n!!}$ for odd $n$:
$I = 2 \times \left( \frac{3-1}{3} \right) = 2 \times \frac{2}{3} = \frac{4}{3}$.
Thus,the value is $\frac{4}{3}$.

(A) Given the equation: $\int\limits_0^x {f\left( t \right)} dt = {x^2} + \int\limits_x^1 {{t^2}f\left( t \right)dt}$.
Differentiating both sides with respect to $x$ using the Leibniz integral rule:
$\frac{d}{dx} \int\limits_0^x {f\left( t \right)} dt = \frac{d}{dx} ({x^2}) + \frac{d}{dx} \int\limits_x^1 {{t^2}f\left( t \right)dt}$.
$f(x) = 2x - x^2 f(x)$.
Rearranging the terms to solve for $f(x)$:
$f(x) + x^2 f(x) = 2x$.
$f(x)(1 + x^2) = 2x$.
$f(x) = \frac{2x}{1 + x^2}$.
Now,find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{(1 + x^2)(2) - (2x)(2x)}{(1 + x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$.
Substitute $x = 1/2$:
$f'(1/2) = \frac{2(1 - (1/2)^2)}{(1 + (1/2)^2)^2} = \frac{2(1 - 1/4)}{(1 + 1/4)^2} = \frac{2(3/4)}{(5/4)^2} = \frac{3/2}{25/16} = \frac{3}{2} \times \frac{16}{25} = \frac{24}{25}$.