Fundamental definite integration Questions in English

(C) Consider the difference $a_{n+1} - a_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2((n+1)x) - \sin^2(nx)}{\sin x} dx$.
Using the identity $\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B)$,we get $\sin^2((n+1)x) - \sin^2(nx) = \sin(x)\sin((2n+1)x)$.
Thus,$a_{n+1} - a_n = \int_0^{\frac{\pi}{2}} \frac{\sin(x)\sin((2n+1)x)}{\sin x} dx = \int_0^{\frac{\pi}{2}} \sin((2n+1)x) dx$.
Evaluating the integral: $a_{n+1} - a_n = \left[ -\frac{\cos((2n+1)x)}{2n+1} \right]_0^{\frac{\pi}{2}} = -\frac{1}{2n+1} (\cos((2n+1)\frac{\pi}{2}) - \cos(0)) = -\frac{1}{2n+1} (0 - 1) = \frac{1}{2n+1}$.
For $n=1$,$a_2 - a_1 = \frac{1}{2(1)+1} = \frac{1}{3}$.
For $n=2$,$a_3 - a_2 = \frac{1}{2(2)+1} = \frac{1}{5}$.
For $n=3$,$a_4 - a_3 = \frac{1}{2(3)+1} = \frac{1}{7}$.
The sequence is $\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \dots$.
Since the reciprocals $3, 5, 7, \dots$ are in Arithmetic Progression,the sequence is in Harmonic Progression.

(D) We are given the integrals $\int_0^1 f(x) dx = 1$,$\int_0^1 x f(x) dx = a$,and $\int_0^1 x^2 f(x) dx = a^2$.
Expanding the integrand $(x-a)^2 f(x) = (x^2 - 2ax + a^2) f(x) = x^2 f(x) - 2ax f(x) + a^2 f(x)$.
Now,integrate term by term:
$\int_0^1 (x-a)^2 f(x) dx = \int_0^1 x^2 f(x) dx - 2a \int_0^1 x f(x) dx + a^2 \int_0^1 f(x) dx$.
Substituting the given values:
$= a^2 - 2a(a) + a^2(1) = a^2 - 2a^2 + a^2 = 0$.

(D) We are given the integral equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$.
Recall the standard integral formula $\int \frac{dt}{|t| \sqrt{t^2-1}} = \sec^{-1}(t) + C$.
Applying the limits of integration,we get:
$\left[ \sec^{-1}(t) \right]_{\sqrt{2}}^x = \frac{\pi}{12}$.
Substituting the limits,we have $\sec^{-1}(x) - \sec^{-1}(\sqrt{2}) = \frac{\pi}{12}$.
Since $\sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$,the equation becomes $\sec^{-1}(x) - \frac{\pi}{4} = \frac{\pi}{12}$.
Adding $\frac{\pi}{4}$ to both sides,we get $\sec^{-1}(x) = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi + 3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.
Therefore,$x = \sec\left(\frac{\pi}{3}\right) = 2$.

(C) Given the integral $\int_3^b \frac{x-1}{2x-x^2} dx = \frac{1}{2}$.
We can rewrite the integrand as $-\frac{1}{2} \int_3^b \frac{-2x+2}{2x-x^2} dx = \frac{1}{2}$.
Multiplying by $-2$,we get $\int_3^b \frac{2-2x}{2x-x^2} dx = -1$.
Integrating,we have $\left[ \ln|2x-x^2| \right]_3^b = -1$.
Substituting the limits: $\ln|2b-b^2| - \ln|6-9| = -1$.
$\ln|2b-b^2| - \ln(3) = -1$.
$\ln|2b-b^2| = \ln(3) - 1$.
Taking the exponential of both sides: $|2b-b^2| = 3e^{-1} = \frac{3}{e}$.
Since $2b-b^2 = -(b^2-2b+1-1) = -(b-1)^2+1$,we have $|1-(b-1)^2| = \frac{3}{e}$.
Assuming $b$ is such that $2b-b^2 < 0$ (as $b > 3$),we have $(b-1)^2 - 1 = \frac{3}{e}$.
Therefore,$(b-1)^2 = 1 + \frac{3}{e}$.

(B) The given limit can be expressed as a Riemann sum:
$S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5} = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{n^5( (r/n)^5 + 1 )} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{(r/n)^4}{(r/n)^5 + 1}$.
By the definition of the definite integral,this is equal to $\int_0^1 \frac{x^4}{x^5+1} dx$.
Let $t = x^5 + 1$,then $dt = 5x^4 dx$,or $x^4 dx = \frac{dt}{5}$.
When $x=0, t=1$. When $x=1, t=2$.
Thus,the integral becomes $\int_1^2 \frac{1}{t} \cdot \frac{dt}{5} = \frac{1}{5} [\log |t|]_1^2 = \frac{1}{5} (\log 2 - \log 1) = \frac{1}{5} \log 2$.

(B) The given integral is of the form of the Beta function,$B(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$.
Here,$m-1 = 5/2 \implies m = 7/2$ and $n-1 = 3/2 \implies n = 5/2$.
So,the integral is $B(7/2, 5/2) = \frac{\Gamma(7/2)\Gamma(5/2)}{\Gamma(7/2 + 5/2)} = \frac{\Gamma(7/2)\Gamma(5/2)}{\Gamma(6)}$.
Using the property $\Gamma(n+1) = n\Gamma(n)$ and $\Gamma(1/2) = \sqrt{\pi}$:
$\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8}$.
$\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$.
$\Gamma(6) = 5! = 120$.
Substituting these values:
Integral $= \frac{(\frac{15\sqrt{\pi}}{8}) \cdot (\frac{3\sqrt{\pi}}{4})}{120} = \frac{45\pi}{32 \cdot 120} = \frac{45\pi}{3840} = \frac{3\pi}{256}$.

(A) Given the integral $\int_{1}^{9} x^2 dx$ with $n = 4$ intervals.
The width of each interval is $h = \frac{9-1}{4} = 2$.
The values of $x$ are $x_0=1, x_1=3, x_2=5, x_3=7, x_4=9$.
The corresponding values of $y = f(x) = x^2$ are:
$y_0 = f(1) = 1^2 = 1$
$y_1 = f(3) = 3^2 = 9$
$y_2 = f(5) = 5^2 = 25$
$y_3 = f(7) = 7^2 = 49$
$y_4 = f(9) = 9^2 = 81$
Using the trapezoidal rule formula:
$\int_{a}^{b} f(x) dx \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4]$
Substituting the values:
$\int_{1}^{9} x^2 dx \approx \frac{2}{2} [1 + 2(9 + 25 + 49) + 81]$
$\int_{1}^{9} x^2 dx \approx 1 [1 + 2(83) + 81]$
$\int_{1}^{9} x^2 dx \approx 1 + 166 + 81 = 248$.

(A) Let $I = \int_1^2 \frac{x^4-1}{x^6-1} d x$.
We simplify the integrand: $\frac{x^4-1}{x^6-1} = \frac{(x^2-1)(x^2+1)}{(x^2-1)(x^4+x^2+1)} = \frac{x^2+1}{x^4+x^2+1}$.
Note that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.
Using partial fractions: $\frac{x^2+1}{(x^2+x+1)(x^2-x+1)} = \frac{1}{2} \left( \frac{1}{x^2+x+1} + \frac{1}{x^2-x+1} \right)$.
Integrating: $I = \frac{1}{2} \int_1^2 \left( \frac{1}{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} + \frac{1}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \right) d x$.
$I = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}\right) + \tan^{-1}\left(\frac{x-1/2}{\sqrt{3}/2}\right) \right]_1^2$.
$I = \frac{1}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right) \right]_1^2$.
Evaluating at limits: $I = \frac{1}{\sqrt{3}} \left[ (\tan^{-1}(\frac{5}{\sqrt{3}}) + \tan^{-1}(\sqrt{3})) - (\tan^{-1}(\sqrt{3}) + \tan^{-1}(\frac{1}{\sqrt{3}})) \right]$.
$I = \frac{1}{\sqrt{3}} \left[ \tan^{-1}(\frac{5}{\sqrt{3}}) - \tan^{-1}(\frac{1}{\sqrt{3}}) \right]$.
Using $\tan^{-1} A - \tan^{-1} B = \tan^{-1}(\frac{A-B}{1+AB})$: $I = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{5/\sqrt{3} - 1/\sqrt{3}}{1 + 5/3} \right) = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{4/\sqrt{3}}{8/3} \right) = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{3}}{2} \right)$.

(C) We need to evaluate the integral $I = \int_{2}^{4} (|x - 2| + |x - 3|) dx$.
Split the integral based on the critical points $x = 2$ and $x = 3$.
For $2 \le x \le 3$,$|x - 2| = x - 2$ and $|x - 3| = 3 - x$.
For $3 \le x \le 4$,$|x - 2| = x - 2$ and $|x - 3| = x - 3$.
Thus,$I = \int_{2}^{3} ((x - 2) + (3 - x)) dx + \int_{3}^{4} ((x - 2) + (x - 3)) dx$.
$I = \int_{2}^{3} (1) dx + \int_{3}^{4} (2x - 5) dx$.
$I = [x]_{2}^{3} + [x^2 - 5x]_{3}^{4}$.
$I = (3 - 2) + ((16 - 20) - (9 - 15))$.
$I = 1 + (-4 - (-6)) = 1 + 2 = 3$.

(D) Let $I = \int_3^5 \sqrt{8x - x^2 - 15} dx$.
We can rewrite the quadratic expression as $-(x^2 - 8x + 15) = -(x^2 - 8x + 16 - 1) = 1 - (x - 4)^2$.
Thus,$I = \int_3^5 \sqrt{1 - (x - 4)^2} dx$.
Let $x - 4 = \sin \theta$,then $dx = \cos \theta d\theta$.
When $x = 3$,$\sin \theta = -1$,so $\theta = -\frac{\pi}{2}$.
When $x = 5$,$\sin \theta = 1$,so $\theta = \frac{\pi}{2}$.
$I = \int_{-\pi/2}^{\pi/2} \sqrt{1 - \sin^2 \theta} \cos \theta d\theta = \int_{-\pi/2}^{\pi/2} \cos^2 \theta d\theta$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get $I = \int_{-\pi/2}^{\pi/2} \frac{1 + \cos 2\theta}{2} d\theta = \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_{-\pi/2}^{\pi/2} = (\frac{\pi}{4} + 0) - (-\frac{\pi}{4} + 0) = \frac{\pi}{2}$.
So,$p = \frac{\pi}{2}$.
Then $\sin p + \operatorname{cosec} p = \sin(\frac{\pi}{2}) + \operatorname{cosec}(\frac{\pi}{2}) = 1 + 1 = 2$.

(D) Given the interval $[0, 6]$ is divided into $n = 6$ equal parts.
The width of each sub-interval is $h = \frac{b - a}{n} = \frac{6 - 0}{6} = 1$.
The values of $f(x) = x^3$ at $x_i = 0, 1, 2, 3, 4, 5, 6$ are:
$y_0 = f(0) = 0$
$y_1 = f(1) = 1$
$y_2 = f(2) = 8$
$y_3 = f(3) = 27$
$y_4 = f(4) = 64$
$y_5 = f(5) = 125$
$y_6 = f(6) = 216$
By the trapezoidal rule:
$\int_0^6 x^3 \, dx \approx \frac{h}{2} \{y_0 + y_6 + 2(y_1 + y_2 + y_3 + y_4 + y_5)\}$
$= \frac{1}{2} \{0 + 216 + 2(1 + 8 + 27 + 64 + 125)\}$
$= \frac{1}{2} \{216 + 2(225)\}$
$= \frac{1}{2} \{216 + 450\} = \frac{666}{2} = 333$.

(C) Given the interval $[2,6]$ is divided into $n=4$ sub-intervals of equal length.
$h = \frac{6-2}{4} = 1$.
Let $f(x) = \frac{1}{x^2-x}$.
The values of $x$ are $x_0=2, x_1=3, x_2=4, x_3=5, x_4=6$.
The corresponding values of $y = f(x)$ are:
$y_0 = f(2) = \frac{1}{4-2} = \frac{1}{2} = 0.5$
$y_1 = f(3) = \frac{1}{9-3} = \frac{1}{6} \approx 0.1667$
$y_2 = f(4) = \frac{1}{16-4} = \frac{1}{12} \approx 0.0833$
$y_3 = f(5) = \frac{1}{25-5} = \frac{1}{20} = 0.05$
$y_4 = f(6) = \frac{1}{36-6} = \frac{1}{30} \approx 0.0333$
Using Simpson's rule: $\int_2^6 f(x) dx \approx \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2y_2]$
$= \frac{1}{3} [\frac{1}{2} + \frac{1}{30} + 4(\frac{1}{6} + \frac{1}{20}) + 2(\frac{1}{12})]$
$= \frac{1}{3} [\frac{16}{30} + 4(\frac{10+3}{60}) + \frac{1}{6}]$
$= \frac{1}{3} [\frac{16}{30} + \frac{52}{60} + \frac{5}{30}] = \frac{1}{3} [\frac{32+52+10}{60}] = \frac{94}{180} = \frac{47}{90} \approx 0.5222$.

(D) Let $I = \int_{1/2}^{1/\sqrt{2}} \frac{1}{(x+\sqrt{1-x^2})(1-x^2)} dx$.
Substitute $x = \sin \theta$,then $dx = \cos \theta d\theta$.
When $x = 1/2$,$\theta = \pi/6$. When $x = 1/\sqrt{2}$,$\theta = \pi/4$.
$I = \int_{\pi/6}^{\pi/4} \frac{\cos \theta d\theta}{(\sin \theta + \cos \theta) \cos^2 \theta} = \int_{\pi/6}^{\pi/4} \frac{d\theta}{(\sin \theta + \cos \theta) \cos \theta}$.
Divide numerator and denominator by $\cos \theta$:
$I = \int_{\pi/6}^{\pi/4} \frac{\sec^2 \theta d\theta}{\tan \theta + 1}$.
Let $u = \tan \theta$,then $du = \sec^2 \theta d\theta$.
When $\theta = \pi/6$,$u = 1/\sqrt{3}$. When $\theta = \pi/4$,$u = 1$.
$I = \int_{1/\sqrt{3}}^{1} \frac{du}{u+1} = [\log |u+1|]_{1/\sqrt{3}}^{1} = \log(2) - \log(1 + 1/\sqrt{3}) = \log \left( \frac{2}{1 + 1/\sqrt{3}} \right) = \log \left( \frac{2\sqrt{3}}{\sqrt{3}+1} \right)$.
Rationalizing the denominator: $\frac{2\sqrt{3}(\sqrt{3}-1)}{3-1} = \sqrt{3}(\sqrt{3}-1) = 3-\sqrt{3}$.
Thus,$I = \log(3-\sqrt{3})$.

(B) Let $I = \int_{-\pi / 4}^{\pi / 3}\left|\tan \left(x-\frac{\pi}{6}\right)\right| d x$.
Let $u = x - \frac{\pi}{6}$,then $du = dx$.
When $x = -\frac{\pi}{4}$,$u = -\frac{\pi}{4} - \frac{\pi}{6} = -\frac{5\pi}{12}$.
When $x = \frac{\pi}{3}$,$u = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
So,$I = \int_{-5\pi/12}^{\pi/6} |\tan u| du = \int_{-5\pi/12}^{0} |\tan u| du + \int_{0}^{\pi/6} |\tan u| du$.
Since $\tan u < 0$ for $u \in [-\frac{5\pi}{12}, 0)$ and $\tan u > 0$ for $u \in (0, \frac{\pi}{6}]$,we have:
$I = \int_{-5\pi/12}^{0} -\tan u du + \int_{0}^{\pi/6} \tan u du$.
$I = [\log |\cos u|]_{-5\pi/12}^{0} + [-\log |\cos u|]_{0}^{\pi/6}$.
$I = (\log 1 - \log |\cos(-5\pi/12)|) - (\log |\cos(\pi/6)| - \log 1)$.
$I = -\log |\cos(5\pi/12)| - \log |\cos(\pi/6)| = -\log |\cos(75^\circ)| - \log |\cos(30^\circ)|$.
Using $\cos(75^\circ) = \cos(45^\circ+30^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos(30^\circ) = \frac{\sqrt{3}}{2}$.
$I = -\log \left( \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{3}}{2} \right) = -\log \left( \frac{3\sqrt{2}-\sqrt{6}}{8} \right) = \log \left( \frac{8}{3\sqrt{2}-\sqrt{6}} \right)$.
Rationalizing the denominator: $\frac{8}{\sqrt{6}(\sqrt{3}-1)} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{8(\sqrt{3}+1)}{\sqrt{6}(3-1)} = \frac{8(\sqrt{3}+1)}{2\sqrt{6}} = \frac{4(\sqrt{3}+1)}{\sqrt{6}} = 2\sqrt{2}(\sqrt{3}+1)$.
Thus,$I = \log(2\sqrt{2}(\sqrt{3}+1))$.

(D) Given the integral $\int_0^4 f(x) d x$ where $f(x) = \frac{1}{1+x^2}$ and $h=1$.
The values of $f(x)$ at $x = 0, 1, 2, 3, 4$ are:
$y_0 = f(0) = 1$
$y_1 = f(1) = \frac{1}{2}$
$y_2 = f(2) = \frac{1}{5}$
$y_3 = f(3) = \frac{1}{10}$
$y_4 = f(4) = \frac{1}{17}$
By using the Trapezoidal rule:
$\int_0^4 f(x) d x = \frac{h}{2} [ (y_0 + y_4) + 2(y_1 + y_2 + y_3) ]$
$= \frac{1}{2} [ (1 + \frac{1}{17}) + 2(\frac{1}{2} + \frac{1}{5} + \frac{1}{10}) ]$
$= \frac{1}{2} [ \frac{18}{17} + 2(\frac{5+2+1}{10}) ]$
$= \frac{1}{2} [ \frac{18}{17} + 2(\frac{8}{10}) ]$
$= \frac{1}{2} [ \frac{18}{17} + \frac{8}{5} ]$
$= \frac{1}{2} [ \frac{90 + 136}{85} ]$
$= \frac{1}{2} [ \frac{226}{85} ] = \frac{113}{85}$

(B) The Trapezoidal rule for numerical integration is given by:
$\text{Distance} = \int_{0}^{10} v \ dt \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3 + y_4) + y_5]$
Here,the interval width $h = 2 \ s$.
The values are $y_0=0, y_1=12, y_2=16, y_3=20, y_4=35, y_5=60$.
Substituting these values into the formula:
$\text{Distance} = \frac{2}{2} [0 + 2(12 + 16 + 20 + 35) + 60]$
$= 1 \times [0 + 2(83) + 60]$
$= 166 + 60$
$= 226 \ \text{m}$.

(B) First,we evaluate the definite integral:
$\int_0^3 (3x^2 - 4x + 2) dx = [x^3 - 2x^2 + 2x]_0^3$
$= (3^3 - 2(3^2) + 2(3)) - (0) = 27 - 18 + 6 = 15$.
Thus,$k = 15$.
Now,we solve the equation $3x^2 - 4x + 2 = \frac{3k}{5}$:
$3x^2 - 4x + 2 = \frac{3(15)}{5} = 9$.
$3x^2 - 4x - 7 = 0$.
Factoring the quadratic equation:
$3x^2 - 7x + 3x - 7 = 0$
$x(3x - 7) + 1(3x - 7) = 0$
$(x + 1)(3x - 7) = 0$.
The roots are $x = -1$ and $x = \frac{7}{3}$.
Since we require the root in the interval $[0, 3]$,the valid root is $x = \frac{7}{3}$.

(D) Let $I = \int_0^{\frac{\pi}{2}} x^3 \sin x \, dx$. Using integration by parts $\int u \, dv = uv - \int v \, du$,let $u = x^3$ and $dv = \sin x \, dx$. Then $du = 3x^2 \, dx$ and $v = -\cos x$.
$I = [-x^3 \cos x]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} 3x^2 \cos x \, dx = 0 + 3 \int_0^{\frac{\pi}{2}} x^2 \cos x \, dx$.
Now,integrate $\int x^2 \cos x \, dx$ by parts again with $u = x^2$ and $dv = \cos x \, dx$. Then $du = 2x \, dx$ and $v = \sin x$.
$I = 3 \left( [x^2 \sin x]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} 2x \sin x \, dx \right) = 3 \left( \frac{\pi^2}{4} - 2 \int_0^{\frac{\pi}{2}} x \sin x \, dx \right)$.
Finally,integrate $\int x \sin x \, dx$ by parts with $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$.
$\int_0^{\frac{\pi}{2}} x \sin x \, dx = [-x \cos x]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \, dx = 0 + [\sin x]_0^{\frac{\pi}{2}} = 1$.
Substituting back: $I = 3 \left( \frac{\pi^2}{4} - 2(1) \right) = \frac{3 \pi^2}{4} - 6$.

(C) Let $I = \int_0^2 \sqrt{(x+3)(2-x)} \, dx$.
Expanding the term inside the square root: $(x+3)(2-x) = -x^2 - x + 6 = 6 - (x^2 + x)$.
Completing the square: $6 - (x^2 + x + \frac{1}{4} - \frac{1}{4}) = \frac{25}{4} - (x + \frac{1}{2})^2$.
So,$I = \int_0^2 \sqrt{(\frac{5}{2})^2 - (x + \frac{1}{2})^2} \, dx$.
Using the formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}(\frac{u}{a}) + C$:
$I = \left[ \frac{2x + 1}{4} \sqrt{(x+3)(2-x)} + \frac{25}{8} \sin^{-1}\left(\frac{2x + 1}{5}\right) \right]_0^2$.
Evaluating at $x=2$: $\frac{5}{4} \sqrt{0} + \frac{25}{8} \sin^{-1}(1) = \frac{25}{8} \cdot \frac{\pi}{2} = \frac{25\pi}{16}$.
Evaluating at $x=0$: $\frac{1}{4} \sqrt{6} + \frac{25}{8} \sin^{-1}(\frac{1}{5})$.
Therefore,$I = \left( \frac{25\pi}{16} \right) - \left( \frac{\sqrt{6}}{4} + \frac{25}{8} \sin^{-1}(\frac{1}{5}) \right) = \frac{25\pi}{16} - \frac{\sqrt{6}}{4} - \frac{25}{8} \sin^{-1}(\frac{1}{5})$.

(C) To evaluate the integral $I = \int_0^{\pi / 4} x^2 \sin 2x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x^2$ and $dv = \sin 2x \, dx$. Then $du = 2x \, dx$ and $v = -\frac{\cos 2x}{2}$.
$I = \left[ -\frac{x^2 \cos 2x}{2} \right]_0^{\pi / 4} - \int_0^{\pi / 4} \left( -\frac{\cos 2x}{2} \right) (2x) \, dx$
$I = \left[ -\frac{x^2 \cos 2x}{2} \right]_0^{\pi / 4} + \int_0^{\pi / 4} x \cos 2x \, dx$
Evaluating the first term: $-\frac{(\pi/4)^2 \cos(\pi/2)}{2} - 0 = 0$ (since $\cos(\pi/2) = 0$).
Now evaluate $\int_0^{\pi / 4} x \cos 2x \, dx$ using integration by parts again:
Let $u = x$ and $dv = \cos 2x \, dx$. Then $du = dx$ and $v = \frac{\sin 2x}{2}$.
$\int_0^{\pi / 4} x \cos 2x \, dx = \left[ \frac{x \sin 2x}{2} \right]_0^{\pi / 4} - \int_0^{\pi / 4} \frac{\sin 2x}{2} \, dx$
$= \left( \frac{(\pi/4) \sin(\pi/2)}{2} - 0 \right) - \left[ -\frac{\cos 2x}{4} \right]_0^{\pi / 4}$
$= \frac{\pi}{8} + \left[ \frac{\cos 2x}{4} \right]_0^{\pi / 4} = \frac{\pi}{8} + \left( \frac{\cos(\pi/2)}{4} - \frac{\cos 0}{4} \right)$
$= \frac{\pi}{8} + (0 - \frac{1}{4}) = \frac{\pi-2}{8}$.

(C) Let $I = \int_0^{\pi / 4} \frac{1}{5 \cos ^2 x+16 \sin ^2 x+8 \sin x \cos x} d x$.
Divide numerator and denominator by $\cos^2 x$:
$I = \int_0^{\pi / 4} \frac{\sec^2 x}{5 + 16 \tan^2 x + 8 \tan x} d x$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
When $x = 0, u = 0$. When $x = \pi/4, u = 1$.
$I = \int_0^1 \frac{du}{16u^2 + 8u + 5} = \int_0^1 \frac{du}{(4u+1)^2 + 4} = \frac{1}{4} \int_0^1 \frac{d(4u+1)}{(4u+1)^2 + 2^2}$.
Using $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{1}{4} \left[ \frac{1}{2} \tan^{-1} \left( \frac{4u+1}{2} \right) \right]_0^1 = \frac{1}{8} [\tan^{-1}(\frac{5}{2}) - \tan^{-1}(\frac{1}{2})]$.
Using $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$:
$I = \frac{1}{8} \tan^{-1} \left( \frac{5/2 - 1/2}{1 + (5/2)(1/2)} \right) = \frac{1}{8} \tan^{-1} \left( \frac{2}{1 + 5/4} \right) = \frac{1}{8} \tan^{-1} \left( \frac{2}{9/4} \right) = \frac{1}{8} \tan^{-1} \left( \frac{8}{9} \right)$.

(NONE) To evaluate the integral $\int_{-2}^1 f(x) dx$,we split the integral at $x=0$ because the definition of $f(x)$ changes at this point.
$\int_{-2}^1 f(x) dx = \int_{-2}^0 (1-2x) dx + \int_0^1 (1+2x) dx$
Now,integrate each part separately:
$\int_{-2}^0 (1-2x) dx = [x - x^2]_{-2}^0 = (0 - 0) - (-2 - (-2)^2) = 0 - (-2 - 4) = 0 - (-6) = 6$
$\int_0^1 (1+2x) dx = [x + x^2]_0^1 = (1 + 1^2) - (0 + 0^2) = 2 - 0 = 2$
Adding these results together:
$\int_{-2}^1 f(x) dx = 6 + 2 = 8$

(C) To solve the integral $I = \int_{-4}^5 \frac{1}{\sqrt{20+x-x^2}} dx$,first complete the square for the quadratic expression inside the square root:
$20+x-x^2 = -(x^2-x-20) = -(x^2-x+\frac{1}{4}-\frac{1}{4}-20) = -((x-\frac{1}{2})^2 - \frac{81}{4}) = \frac{81}{4} - (x-\frac{1}{2})^2$.
Now the integral becomes $I = \int_{-4}^5 \frac{1}{\sqrt{(\frac{9}{2})^2 - (x-\frac{1}{2})^2}} dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^2-u^2}} du = \sin^{-1}(\frac{u}{a}) + C$,we get:
$I = [\sin^{-1}(\frac{x-1/2}{9/2})]_{-4}^5 = [\sin^{-1}(\frac{2x-1}{9})]_{-4}^5$.
Evaluating at the limits:
$I = \sin^{-1}(\frac{2(5)-1}{9}) - \sin^{-1}(\frac{2(-4)-1}{9}) = \sin^{-1}(\frac{9}{9}) - \sin^{-1}(\frac{-9}{9}) = \sin^{-1}(1) - \sin^{-1}(-1)$.
Since $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(-1) = -\frac{\pi}{2}$,we have:
$I = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

(D) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{\cos x - \sqrt{3} \sin x}$.
Multiply and divide by $2$ to rewrite the denominator:
$I = \int_0^{\frac{\pi}{2}} \frac{dx}{2(\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x)} = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{dx}{\sin(\frac{\pi}{6} - x)}$.
Using the identity $\sin(\theta) = \cos(\frac{\pi}{2} - \theta)$,we have $\sin(\frac{\pi}{6} - x) = \cos(\frac{\pi}{2} - (\frac{\pi}{6} - x)) = \cos(x + \frac{\pi}{3})$.
Thus,$I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sec(x + \frac{\pi}{3}) dx$.
The integral of $\sec(u)$ is $\log|\sec u + \tan u|$.
$I = \frac{1}{2} [\log|\sec(x + \frac{\pi}{3}) + \tan(x + \frac{\pi}{3})|]_0^{\frac{\pi}{2}}$.
Evaluating at the limits:
At $x = \frac{\pi}{2}$,$\sec(\frac{5\pi}{6}) = -\frac{2}{\sqrt{3}}$ and $\tan(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$.
At $x = 0$,$\sec(\frac{\pi}{3}) = 2$ and $\tan(\frac{\pi}{3}) = \sqrt{3}$.
$I = \frac{1}{2} [\log|-\frac{3}{\sqrt{3}}| - \log|2 + \sqrt{3}|] = \frac{1}{2} [\log(\sqrt{3}) - \log(2 + \sqrt{3})]$.
This simplifies to $\frac{1}{2} \log(\frac{\sqrt{3}}{2+\sqrt{3}})$.
Rationalizing the denominator: $\frac{\sqrt{3}(2-\sqrt{3})}{4-3} = 2\sqrt{3} - 3$.
Thus,$I = \frac{1}{2} \log(2\sqrt{3} - 3)$.

(C) Let $I = \int_2^5 \sqrt{\frac{5-x}{x-2}} \, dx$. Put $x = 2 \cos^2 \theta + 5 \sin^2 \theta = 2 + 3 \sin^2 \theta$.
Then $dx = 6 \sin \theta \cos \theta \, d\theta$.
When $x=2, \theta=0$. When $x=5, \theta=\frac{\pi}{2}$.
$I = \int_0^{\pi/2} \sqrt{\frac{5-(2+3\sin^2 \theta)}{(2+3\sin^2 \theta)-2}} \cdot 6 \sin \theta \cos \theta \, d\theta$
$I = \int_0^{\pi/2} \sqrt{\frac{3-3\sin^2 \theta}{3\sin^2 \theta}} \cdot 6 \sin \theta \cos \theta \, d\theta$
$I = \int_0^{\pi/2} \frac{\cos \theta}{\sin \theta} \cdot 6 \sin \theta \cos \theta \, d\theta = 6 \int_0^{\pi/2} \cos^2 \theta \, d\theta$
$I = 6 \int_0^{\pi/2} \frac{1+\cos 2\theta}{2} \, d\theta = 3 \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\pi/2} = 3 \left( \frac{\pi}{2} + 0 - 0 \right) = \frac{3\pi}{2}$.

(A) Let $I = \int_0^{\pi / 4} \frac{\sec x}{1+2 \sin ^2 x} d x$.
Multiply numerator and denominator by $\cos x$:
$I = \int_0^{\pi / 4} \frac{\cos x}{\cos ^2 x(1+2 \sin ^2 x)} d x = \int_0^{\pi / 4} \frac{\cos x}{(1-\sin ^2 x)(1+2 \sin ^2 x)} d x$.
Let $\sin x = t$,then $\cos x d x = d t$.
When $x = 0, t = 0$. When $x = \frac{\pi}{4}, t = \frac{1}{\sqrt{2}}$.
$I = \int_0^{1 / \sqrt{2}} \frac{d t}{(1-t^2)(1+2 t^2)}$.
Using partial fractions: $\frac{1}{(1-t^2)(1+2 t^2)} = \frac{1}{3} \left( \frac{1}{1-t^2} + \frac{2}{1+2 t^2} \right)$.
$I = \frac{1}{3} \int_0^{1 / \sqrt{2}} \left( \frac{1}{1-t^2} + \frac{2}{1+2 t^2} \right) d t$.
$I = \frac{1}{3} \left[ \frac{1}{2} \log \left| \frac{1+t}{1-t} \right| + \frac{2}{\sqrt{2}} \tan ^{-1} (\sqrt{2} t) \right]_0^{1 / \sqrt{2}}$.
$I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{1+1/\sqrt{2}}{1-1/\sqrt{2}} \right) + \sqrt{2} \tan ^{-1} (1) \right]$.
$I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right) + \sqrt{2} \cdot \frac{\pi}{4} \right]$.
Since $\frac{\sqrt{2}+1}{\sqrt{2}-1} = (\sqrt{2}+1)^2$,we have $\log (\sqrt{2}+1)^2 = 2 \log (\sqrt{2}+1)$.
$I = \frac{1}{3} \left[ \log (\sqrt{2}+1) + \frac{\pi \sqrt{2}}{4} \right] = \frac{1}{3} \log (\sqrt{2}+1) + \frac{\pi \sqrt{2}}{12}$.

(C) Let $I = \int_0^2 \frac{x}{(2-x)^{3/4}} dx$.
Substitute $t = 2-x$,then $dx = -dt$.
When $x=0, t=2$ and when $x=2, t=0$.
$I = \int_2^0 \frac{2-t}{t^{3/4}} (-dt) = \int_0^2 \frac{2-t}{t^{3/4}} dt$.
$I = \int_0^2 (2t^{-3/4} - t^{1/4}) dt$.
$I = [2 \cdot \frac{t^{1/4}}{1/4} - \frac{t^{5/4}}{5/4}]_0^2$.
$I = [8t^{1/4} - \frac{4}{5}t^{5/4}]_0^2$.
$I = 8(2^{1/4}) - \frac{4}{5}(2^{5/4}) = 8(2^{1/4}) - \frac{4}{5}(2 \cdot 2^{1/4})$.
$I = 2^{1/4} (8 - \frac{8}{5}) = 2^{1/4} (\frac{40-8}{5}) = \frac{32}{5} 2^{1/4}$.

(D) Let $I = \int_0^3 \left(\sin \left(\frac{\pi}{3} x\right) - \cos \left(\frac{\pi}{3} x\right)\right) dx$.
Integrating term by term:
$I = \int_0^3 \sin \left(\frac{\pi}{3} x\right) dx - \int_0^3 \cos \left(\frac{\pi}{3} x\right) dx$.
Using the formula $\int \sin(ax) dx = -\frac{1}{a} \cos(ax)$ and $\int \cos(ax) dx = \frac{1}{a} \sin(ax)$:
$I = \left[ -\frac{3}{\pi} \cos \left(\frac{\pi}{3} x\right) \right]_0^3 - \left[ \frac{3}{\pi} \sin \left(\frac{\pi}{3} x\right) \right]_0^3$.
$I = -\frac{3}{\pi} [\cos(\pi) - \cos(0)] - \frac{3}{\pi} [\sin(\pi) - \sin(0)]$.
$I = -\frac{3}{\pi} [-1 - 1] - \frac{3}{\pi} [0 - 0]$.
$I = -\frac{3}{\pi} [-2] = \frac{6}{\pi}$.

(B) We need to evaluate the definite integral $\int_0^2 |1-x^2| dx$.
First,we analyze the absolute value function $|1-x^2|$. The expression $1-x^2$ changes sign at $x=1$ within the interval $[0, 2]$.
For $0 \leq x < 1$,$1-x^2 \geq 0$,so $|1-x^2| = 1-x^2$.
For $1 \leq x \leq 2$,$1-x^2 \leq 0$,so $|1-x^2| = -(1-x^2) = x^2-1$.
Using the property of definite integrals,we split the integral at $x=1$:
$\int_0^2 |1-x^2| dx = \int_0^1 (1-x^2) dx + \int_1^2 (x^2-1) dx$
Now,we integrate each part:
$\int_0^1 (1-x^2) dx = [x - \frac{x^3}{3}]_0^1 = (1 - \frac{1}{3}) - 0 = \frac{2}{3}$
$\int_1^2 (x^2-1) dx = [\frac{x^3}{3} - x]_1^2 = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{2}{3} - (-\frac{2}{3}) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$
Adding these results together:
$\frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2$

(B) The fractional part function $\{x\}$ is defined as $\{x\} = x - \lfloor x \rfloor$.
For the interval $[0, 2)$,we have:
$\{x\} = x$ for $0 \leq x < 1$
$\{x\} = x - 1$ for $1 \leq x < 2$
Therefore,the integral can be split as:
$\int_0^2 \{x\} dx = \int_0^1 x dx + \int_1^2 (x - 1) dx$
Evaluating the first part: $\int_0^1 x dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$
Evaluating the second part: $\int_1^2 (x - 1) dx$. Let $u = x - 1$,then $du = dx$. When $x=1, u=0$; when $x=2, u=1$.
$\int_0^1 u du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2}$
Adding both parts: $\frac{1}{2} + \frac{1}{2} = 1$.

(C) Let $I = \int_0^{\pi / 2} \frac{\pi \sin x}{1+\cos ^2 x} d x$.
Substitute $\cos x = t$,then $-\sin x d x = d t$,or $\sin x d x = -d t$.
When $x = 0$,$t = \cos(0) = 1$.
When $x = \pi / 2$,$t = \cos(\pi / 2) = 0$.
The integral becomes:
$I = \pi \int_1^0 \frac{-d t}{1+t^2} = \pi \int_0^1 \frac{d t}{1+t^2}$.
Using the standard integral $\int \frac{1}{1+t^2} d t = \tan^{-1}(t)$:
$I = \pi [\tan^{-1}(t)]_0^1 = \pi (\tan^{-1}(1) - \tan^{-1}(0))$.
Since $\tan^{-1}(1) = \pi / 4$ and $\tan^{-1}(0) = 0$:
$I = \pi (\pi / 4 - 0) = \frac{\pi^2}{4}$.

(B) Let $I = \int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$.
Using the substitution $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$,we have:
$I = \int_0^{\pi / 2} \frac{dx}{4 + 5 \left( \frac{2 \tan(x/2)}{1+\tan^2(x/2)} \right)} = \int_0^{\pi / 2} \frac{\sec^2(x/2) dx}{4(1+\tan^2(x/2)) + 10 \tan(x/2)} = \int_0^{\pi / 2} \frac{\sec^2(x/2) dx}{4 \tan^2(x/2) + 10 \tan(x/2) + 4}$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dt$.
When $x=0, t=0$; when $x=\pi/2, t=1$.
$I = \int_0^1 \frac{2 dt}{4t^2 + 10t + 4} = \frac{1}{2} \int_0^1 \frac{dt}{t^2 + \frac{5}{2}t + 1}$.
Completing the square: $t^2 + \frac{5}{2}t + 1 = (t + \frac{5}{4})^2 - \frac{25}{16} + 1 = (t + \frac{5}{4})^2 - (\frac{3}{4})^2$.
$I = \frac{1}{2} \int_0^1 \frac{dt}{(t + \frac{5}{4})^2 - (\frac{3}{4})^2} = \frac{1}{2} \cdot \frac{1}{2(3/4)} \ln \left| \frac{t + 5/4 - 3/4}{t + 5/4 + 3/4} \right|_0^1$.
$I = \frac{1}{3} \left[ \ln \left| \frac{t + 1/2}{t + 2} \right| \right]_0^1 = \frac{1}{3} \left[ \ln \left( \frac{3/2}{3} \right) - \ln \left( \frac{1/2}{2} \right) \right]$.
$I = \frac{1}{3} [ \ln(1/2) - \ln(1/4) ] = \frac{1}{3} \ln \left( \frac{1/2}{1/4} \right) = \frac{1}{3} \ln 2$.

(B) Given,$f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x = (\sin ^3 x+\cos ^3 x)^2$.
We need to evaluate $I=\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)} d x$.
Using $\sin 2x = 2 \sin x \cos x$,we have $\sin ^2 2x = 4 \sin ^2 x \cos ^2 x$.
Thus,$I = \int_0^{\pi / 4} \frac{4 \sin ^2 x \cos ^2 x}{(\sin ^3 x+\cos ^3 x)^2} d x$.
Dividing numerator and denominator by $\cos ^6 x$,we get:
$I = \int_0^{\pi / 4} \frac{4 \tan ^2 x \sec ^2 x}{(\tan ^3 x+1)^2} d x$.
Let $t = \tan ^3 x$,then $dt = 3 \tan ^2 x \sec ^2 x dx$,which implies $\tan ^2 x \sec ^2 x dx = \frac{dt}{3}$.
When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$.
Substituting these into the integral:
$I = \int_0^1 \frac{4 \cdot (dt/3)}{(t+1)^2} = \frac{4}{3} \int_0^1 (t+1)^{-2} dt$.
$I = \frac{4}{3} \left[ \frac{-1}{t+1} \right]_0^1 = \frac{4}{3} \left( -\frac{1}{2} - (-1) \right) = \frac{4}{3} \left( \frac{1}{2} \right) = \frac{2}{3}$.

(A) Let $I = \int_{\pi/4}^{\pi/2} \cot^9 x \, dx$.
Substitute $t = \sin x$,so $dt = \cos x \, dx$.
Since $\cot^9 x = \frac{\cos^9 x}{\sin^9 x} = \frac{(\cos^2 x)^4 \cos x}{\sin^9 x} = \frac{(1-t^2)^4}{t^9}$,we have:
$I = \int_{1/\sqrt{2}}^{1} \frac{(1-t^2)^4}{t^9} \, dt$.
Expanding $(1-t^2)^4 = 1 - 4t^2 + 6t^4 - 4t^6 + t^8$:
$I = \int_{1/\sqrt{2}}^{1} (t^{-9} - 4t^{-7} + 6t^{-5} - 4t^{-3} + t^{-1}) \, dt$.
Integrating term by term:
$I = \left[ \frac{t^{-8}}{-8} - 4\frac{t^{-6}}{-6} + 6\frac{t^{-4}}{-4} - 4\frac{t^{-2}}{-2} + \log|t| \right]_{1/\sqrt{2}}^{1}$.
$I = \left[ -\frac{1}{8t^8} + \frac{2}{3t^6} - \frac{3}{2t^4} + \frac{2}{t^2} + \log t \right]_{1/\sqrt{2}}^{1}$.
Evaluating at the limits:
At $t=1$: $-\frac{1}{8} + \frac{2}{3} - \frac{3}{2} + 2 + \log 1 = \frac{-3 + 16 - 36 + 48}{24} = \frac{25}{24}$.
At $t=1/\sqrt{2}$: $-\frac{1}{8(1/16)} + \frac{2}{3(1/8)} - \frac{3}{2(1/4)} + \frac{2}{1/2} + \log(1/\sqrt{2}) = -2 + \frac{16}{3} - 6 + 4 - \frac{1}{2} \log 2 = \frac{16}{3} - 4 - \frac{1}{2} \log 2 = \frac{4}{3} - \frac{1}{2} \log 2 = \frac{32}{24} - \frac{1}{2} \log 2$.
$I = \frac{25}{24} - (\frac{32}{24} - \frac{1}{2} \log 2) = -\frac{7}{24} + \frac{1}{2} \log 2$.

(C) Given,$I=\int_0^{\pi / 2} \frac{d x}{5+3 \sin x}=\lambda \tan ^{-1}\left(\frac{1}{2}\right)$.
Using the substitution $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$,we have:
$I = \int_0^{\pi / 2} \frac{dx}{5 + 3 \left( \frac{2 \tan(x/2)}{1+\tan^2(x/2)} \right)} = \int_0^{\pi / 2} \frac{(1+\tan^2(x/2)) dx}{5 + 5\tan^2(x/2) + 6\tan(x/2)} = \int_0^{\pi / 2} \frac{\sec^2(x/2) dx}{5\tan^2(x/2) + 6\tan(x/2) + 5}$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dt$.
When $x=0, t=0$; when $x=\pi/2, t=1$.
$I = \int_0^1 \frac{2 dt}{5t^2 + 6t + 5} = \frac{2}{5} \int_0^1 \frac{dt}{t^2 + \frac{6}{5}t + 1} = \frac{2}{5} \int_0^1 \frac{dt}{(t + 3/5)^2 + 1 - 9/25} = \frac{2}{5} \int_0^1 \frac{dt}{(t + 3/5)^2 + (4/5)^2}$.
Using $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})$:
$I = \frac{2}{5} \cdot \frac{5}{4} \left[ \tan^{-1} \left( \frac{t + 3/5}{4/5} \right) \right]_0^1 = \frac{1}{2} \left[ \tan^{-1} \left( \frac{5t+3}{4} \right) \right]_0^1 = \frac{1}{2} \left( \tan^{-1}(2) - \tan^{-1}(3/4) \right)$.
Using $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$:
$I = \frac{1}{2} \tan^{-1} \left( \frac{2 - 3/4}{1 + 2(3/4)} \right) = \frac{1}{2} \tan^{-1} \left( \frac{5/4}{1 + 3/2} \right) = \frac{1}{2} \tan^{-1} \left( \frac{5/4}{5/2} \right) = \frac{1}{2} \tan^{-1} \left( \frac{1}{2} \right)$.
Comparing with $\lambda \tan^{-1}(1/2)$,we get $\lambda = 1/2$.

(A) Let $I = \int_0^{\pi / 4} (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$.
$I = \int_0^{\pi / 4} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right) \, dx$.
$I = \int_0^{\pi / 4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$.
Multiply numerator and denominator by $\sqrt{2}$:
$I = \int_0^{\pi / 4} \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi / 4} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$.
Let $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) \, dx$.
When $x = 0$,$t = -1$. When $x = \pi/4$,$t = 0$.
$I = \sqrt{2} \int_{-1}^0 \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^0$.
$I = \sqrt{2} [\sin^{-1}(0) - \sin^{-1}(-1)] = \sqrt{2} [0 - (-\pi/2)] = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.

(D) Let $I = \int_0^{\pi / 4} \frac{\sin x+\cos x}{7+9 \sin 2 x} d x$.
We know that $(\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x$.
Therefore,$\sin 2x = 1 - (\sin x - \cos x)^2$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi/4$,$t = \sin(\pi/4) - \cos(\pi/4) = 0$.
Substituting these into the integral:
$I = \int_{-1}^0 \frac{dt}{7 + 9(1 - t^2)} = \int_{-1}^0 \frac{dt}{16 - 9t^2}$.
$I = \frac{1}{9} \int_{-1}^0 \frac{dt}{(4/3)^2 - t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$,we get:
$I = \frac{1}{9} \cdot \frac{1}{2(4/3)} \left[ \log \left| \frac{4/3 + t}{4/3 - t} \right| \right]_{-1}^0 = \frac{1}{24} \left[ \log \left| \frac{4+3t}{4-3t} \right| \right]_{-1}^0$.
$I = \frac{1}{24} [\log(1) - \log| (4-3)/(4+3) |] = \frac{1}{24} [0 - \log(1/7)] = \frac{1}{24} \log 7$.

(A) Let $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$.
Rationalizing the integrand,we get:
$I = \int_0^1 \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} \, dx = \int_0^1 \frac{1-x}{\sqrt{1-x^2}} \, dx$.
Splitting the integral:
$I = \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx - \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx$.
Evaluating the first part: $\int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx = [\sin^{-1} x]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.
Evaluating the second part: Let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$.
$\int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int_1^0 u^{-1/2} \, du = \frac{1}{2} \int_0^1 u^{-1/2} \, du = \frac{1}{2} [2\sqrt{u}]_0^1 = [\sqrt{u}]_0^1 = 1 - 0 = 1$.
Thus,$I = \frac{\pi}{2} - 1$.

(B) Let $I = \int_0^\pi \frac{1}{1+\sin x} dx$.
Using the identity $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$,we get:
$I = \int_0^\pi \frac{1}{1+\frac{2 \tan(x/2)}{1+\tan^2(x/2)}} dx = \int_0^\pi \frac{1+\tan^2(x/2)}{1+\tan^2(x/2)+2 \tan(x/2)} dx$.
Since $1+\tan^2(x/2) = \sec^2(x/2)$,we have:
$I = \int_0^\pi \frac{\sec^2(x/2)}{(1+\tan(x/2))^2} dx$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,which implies $\sec^2(x/2) dx = 2 dt$.
As $x \to 0$,$t \to 0$. As $x \to \pi$,$t \to \infty$.
Thus,$I = \int_0^\infty \frac{2 dt}{(1+t)^2} = 2 \left[ -\frac{1}{1+t} \right]_0^\infty = 2(0 - (-1)) = 2$.

(A) We need to evaluate the integral $I = \int_0^3 \frac{3x+1}{x^2+9} dx$.
Split the integral into two parts:
$I = \int_0^3 \frac{3x}{x^2+9} dx + \int_0^3 \frac{1}{x^2+9} dx$
For the first part,let $u = x^2+9$,then $du = 2x dx$,so $x dx = \frac{du}{2}$.
$\int_0^3 \frac{3x}{x^2+9} dx = \frac{3}{2} \int_9^{18} \frac{du}{u} = \frac{3}{2} [\log |u|]_9^{18} = \frac{3}{2} (\log 18 - \log 9) = \frac{3}{2} \log 2 = \log (2^{3/2}) = \log (2\sqrt{2})$.
For the second part,use the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$\int_0^3 \frac{1}{x^2+3^2} dx = [\frac{1}{3} \tan^{-1}(\frac{x}{3})]_0^3 = \frac{1}{3} (\tan^{-1}(1) - \tan^{-1}(0)) = \frac{1}{3} (\frac{\pi}{4} - 0) = \frac{\pi}{12}$.
Adding both parts,we get $I = \log (2\sqrt{2}) + \frac{\pi}{12}$.

(B) Let $I = \int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$.
Put $x = \cos \theta$,then $d x = -\sin \theta d \theta$.
When $x = 0$,$\theta = \frac{\pi}{2}$ and when $x = 1$,$\theta = 0$.
Substituting these into the integral:
$I = \int_{\pi/2}^0 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right) (-\sin \theta) d \theta$.
Using the identity $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}} = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} = \cot(\theta/2) = \tan(\frac{\pi}{2} - \frac{\theta}{2})$.
$I = \int_0^{\pi/2} \sin \left(2 \tan ^{-1} \tan \left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) \sin \theta d \theta$.
$I = \int_0^{\pi/2} \sin \left(\pi - \theta\right) \sin \theta d \theta = \int_0^{\pi/2} \sin^2 \theta d \theta$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:
$I = \int_0^{\pi/2} \frac{1 - \cos 2\theta}{2} d \theta = \frac{1}{2} \left[ \theta - \frac{\sin 2\theta}{2} \right]_0^{\pi/2}$.
$I = \frac{1}{2} \left[ (\frac{\pi}{2} - 0) - (0 - 0) \right] = \frac{\pi}{4}$.