Fundamental definite integration Questions in English

(A) Let $I = \int_0^{\pi /4} \sec x \log (\sec x + \tan x) \, dx$.
Substitute $t = \log (\sec x + \tan x)$.
Then $dt = \frac{1}{\sec x + \tan x} (\sec x \tan x + \sec^2 x) \, dx = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \, dx = \sec x \, dx$.
Change the limits of integration:
When $x = 0$,$t = \log (\sec 0 + \tan 0) = \log (1 + 0) = \log 1 = 0$.
When $x = \pi / 4$,$t = \log (\sec(\pi / 4) + \tan(\pi / 4)) = \log (\sqrt{2} + 1)$.
Now,the integral becomes:
$I = \int_0^{\log (\sqrt{2} + 1)} t \, dt = \left[ \frac{t^2}{2} \right]_0^{\log (\sqrt{2} + 1)} = \frac{1}{2} [\log (\sqrt{2} + 1)]^2$.
Thus,the correct option is $A$.

(B) To evaluate the integral $I = \int_0^{2/3} \frac{dx}{4 + 9x^2}$,we first rewrite the denominator:
$I = \int_0^{2/3} \frac{dx}{2^2 + (3x)^2}$.
Let $3x = u$,then $3dx = du$,which implies $dx = \frac{du}{3}$.
When $x = 0$,$u = 0$. When $x = 2/3$,$u = 2$.
Substituting these into the integral:
$I = \int_0^2 \frac{1}{2^2 + u^2} \cdot \frac{du}{3} = \frac{1}{3} \int_0^2 \frac{du}{2^2 + u^2}$.
Using the standard formula $\int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C$:
$I = \frac{1}{3} \left[ \frac{1}{2} \tan^{-1}(\frac{u}{2}) \right]_0^2 = \frac{1}{6} [\tan^{-1}(1) - \tan^{-1}(0)]$.
Since $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(0) = 0$:
$I = \frac{1}{6} \times \frac{\pi}{4} = \frac{\pi}{24}$.

(A) Let $I = \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$.
We can rewrite the integrand as $\frac{x^4 - 1 + 2}{x^2 + 1} = \frac{(x^2 - 1)(x^2 + 1) + 2}{x^2 + 1} = x^2 - 1 + \frac{2}{x^2 + 1}$.
Now,integrate term by term:
$I = \int_0^1 (x^2 - 1) \, dx + 2 \int_0^1 \frac{1}{x^2 + 1} \, dx$.
Evaluating the integrals:
$I = \left[ \frac{x^3}{3} - x \right]_0^1 + 2 \left[ \tan^{-1} x \right]_0^1$.
$I = \left( \frac{1}{3} - 1 \right) - (0) + 2 \left( \tan^{-1}(1) - \tan^{-1}(0) \right)$.
$I = -\frac{2}{3} + 2 \left( \frac{\pi}{4} - 0 \right)$.
$I = -\frac{2}{3} + \frac{\pi}{2} = \frac{3\pi - 4}{6}$.

(C) Let $I = \int_0^{\pi /4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$.
We can rewrite the integrand as:
$I = \int_0^{\pi /4} \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) dx = \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$.
Multiply the numerator and denominator by $\sqrt{2}$:
$I = \sqrt{2} \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) \, dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi / 4$,$t = \sin(\pi / 4) - \cos(\pi / 4) = 0$.
Substituting these into the integral:
$I = \sqrt{2} \int_{-1}^0 \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^0$.
$I = \sqrt{2} [\sin^{-1}(0) - \sin^{-1}(-1)] = \sqrt{2} [0 - (-\pi / 2)] = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.

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

(C) Let $I = \int_1^e \frac{1}{x} \, dx$.
We know that the antiderivative of $\frac{1}{x}$ is $\ln|x|$.
Applying the Fundamental Theorem of Calculus:
$I = [\ln|x|]_1^e$
$I = \ln(e) - \ln(1)$
Since $\ln(e) = 1$ and $\ln(1) = 0$,
$I = 1 - 0 = 1$.
Thus,the correct option is $C$.

(A) Given integral: $I = \int_{1}^{x} \frac{\log(x^2)}{x} \, dx$
Using the property $\log(a^b) = b \log a$,we have:
$I = \int_{1}^{x} \frac{2 \log x}{x} \, dx$
Let $t = \log x$,then $dt = \frac{1}{x} \, dx$.
When $x = 1$,$t = \log(1) = 0$.
When $x = x$,$t = \log x$.
Substituting these into the integral:
$I = \int_{0}^{\log x} 2t \, dt$
$I = 2 \left[ \frac{t^2}{2} \right]_{0}^{\log x}$
$I = [t^2]_{0}^{\log x}$
$I = (\log x)^2 - 0^2 = (\log x)^2$
Thus,the correct option is $A$.

(D) Let $I = \int_0^{\pi /2} {\frac{{dx}}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}.} $
Dividing the numerator and denominator by ${\cos ^2}x$,we get
$I = \int_0^{\pi /2} {\frac{{\sec^2 x \, dx}}{{{a^2} + {b^2}\tan^2 x}}} $
Substituting $b \tan x = t$,we have $b \sec^2 x \, dx = dt$,or $\sec^2 x \, dx = \frac{dt}{b}$.
When $x = 0$,$t = 0$. When $x = \frac{\pi}{2}$,$t \to \infty$.
Therefore,$I = \int_0^\infty {\frac{{dt/b}}{{{a^2} + {t^2}}}} = \frac{1}{b} \int_0^\infty \frac{dt}{a^2 + t^2}$
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get
$I = \frac{1}{b} \left[ \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) \right]_0^\infty = \frac{1}{ab} \left( \tan^{-1}(\infty) - \tan^{-1}(0) \right)$
$I = \frac{1}{ab} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{2ab}$.

(D) Let $I = \int_0^{\pi /4} (\cos x - \sin x) dx + \int_{\pi /4}^{5\pi /4} (\sin x - \cos x) dx + \int_{2\pi }^{\pi /4} (\cos x - \sin x) dx$.
Evaluating each integral:
$1$. $\int_0^{\pi /4} (\cos x - \sin x) dx = [\sin x + \cos x]_0^{\pi /4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.
$2$. $\int_{\pi /4}^{5\pi /4} (\sin x - \cos x) dx = [-\cos x - \sin x]_{\pi /4}^{5\pi /4} = -[\cos x + \sin x]_{\pi /4}^{5\pi /4} = -[(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})] = -[-\sqrt{2} - \sqrt{2}] = 2\sqrt{2}$.
$3$. $\int_{2\pi }^{\pi /4} (\cos x - \sin x) dx = [\sin x + \cos x]_{2\pi }^{\pi /4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.
Summing these values: $I = (\sqrt{2} - 1) + 2\sqrt{2} + (\sqrt{2} - 1) = 4\sqrt{2} - 2$.

(B) Given the inequality: $\int_{0}^{a} x \, dx \le a + 4$
Evaluating the definite integral: $\left[ \frac{x^2}{2} \right]_{0}^{a} \le a + 4$
$\frac{a^2}{2} \le a + 4$
Multiplying by $2$: $a^2 \le 2a + 8$
Rearranging the terms: $a^2 - 2a - 8 \le 0$
Factoring the quadratic expression: $(a - 4)(a + 2) \le 0$
For the product to be less than or equal to zero,$a$ must lie between the roots: $-2 \le a \le 4$.

(B) Let $I = \int_{-1}^{0} \frac{dx}{x^2 + 2x + 2}$.
First,complete the square in the denominator: $x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1^2$.
Thus,the integral becomes $I = \int_{-1}^{0} \frac{dx}{(x + 1)^2 + 1^2}$.
Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = [\tan^{-1}(x + 1)]_{-1}^{0}$.
Evaluating the limits:
$I = \tan^{-1}(0 + 1) - \tan^{-1}(-1 + 1) = \tan^{-1}(1) - \tan^{-1}(0)$.
Since $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(0) = 0$,we have:
$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

(A) We know that the integral of $\frac{1}{1 + x^2}$ with respect to $x$ is $\tan^{-1}(x)$.
Applying the fundamental theorem of calculus for the definite integral:
$\int_1^{\sqrt{3}} \frac{1}{1 + x^2} dx = [\tan^{-1}(x)]_1^{\sqrt{3}}$
$= \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)$
Since $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ and $\tan^{-1}(1) = \frac{\pi}{4}$,we have:
$= \frac{\pi}{3} - \frac{\pi}{4}$
$= \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$.

(D) Let $I = \int_{1}^{3} (x - 1)(x - 2)(x - 3) \, dx$.
Expanding the integrand: $(x - 1)(x - 2)(x - 3) = (x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 - 3x^2 + 9x + 2x - 6 = x^3 - 6x^2 + 11x - 6$.
Now,integrate term by term:
$I = \left[ \frac{x^4}{4} - \frac{6x^3}{3} + \frac{11x^2}{2} - 6x \right]_{1}^{3} = \left[ \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x \right]_{1}^{3}$.
Evaluating at the upper limit $x = 3$:
$\left( \frac{81}{4} - 2(27) + \frac{11(9)}{2} - 6(3) \right) = \frac{81}{4} - 54 + \frac{99}{2} - 18 = \frac{81 + 198}{4} - 72 = \frac{279}{4} - 72 = \frac{279 - 288}{4} = -\frac{9}{4}$.
Evaluating at the lower limit $x = 1$:
$\left( \frac{1}{4} - 2(1) + \frac{11}{2} - 6(1) \right) = \frac{1}{4} - 2 + \frac{11}{2} - 6 = \frac{1 + 22}{4} - 8 = \frac{23}{4} - 8 = \frac{23 - 32}{4} = -\frac{9}{4}$.
Therefore,$I = -\frac{9}{4} - (-\frac{9}{4}) = 0$.

(C) Let $I = \int_{2}^{3} \frac{dx}{x(x - 1)}$.
Using partial fractions,we can write $\frac{1}{x(x - 1)} = \frac{1}{x - 1} - \frac{1}{x}$.
Therefore,$I = \int_{2}^{3} \left( \frac{1}{x - 1} - \frac{1}{x} \right) dx$.
Integrating term by term,we get $I = [\log|x - 1| - \log|x|]_{2}^{3}$.
Using the property $\log a - \log b = \log(a/b)$,we have $I = [\log|\frac{x - 1}{x}|]_{2}^{3}$.
Substituting the limits: $I = \log|\frac{3 - 1}{3}| - \log|\frac{2 - 1}{2}|$.
$I = \log(2/3) - \log(1/2) = \log(\frac{2/3}{1/2}) = \log(4/3)$.

(C) Let $I = \int_{0}^{\pi} |\sin^3 \theta| \, d\theta$.
Since $\sin \theta \ge 0$ for all $\theta \in [0, \pi]$,we have $|\sin^3 \theta| = \sin^3 \theta$.
Therefore,$I = \int_{0}^{\pi} \sin^3 \theta \, d\theta$.
Using the identity $\sin^3 \theta = \sin \theta (1 - \cos^2 \theta)$,we get:
$I = \int_{0}^{\pi} \sin \theta (1 - \cos^2 \theta) \, d\theta = \int_{0}^{\pi} \sin \theta \, d\theta - \int_{0}^{\pi} \sin \theta \cos^2 \theta \, d\theta$.
Evaluating the integrals:
$\int_{0}^{\pi} \sin \theta \, d\theta = [-\cos \theta]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$.
For the second part,let $u = \cos \theta$,then $du = -\sin \theta \, d\theta$.
When $\theta = 0, u = 1$; when $\theta = \pi, u = -1$.
$-\int_{0}^{\pi} \sin \theta \cos^2 \theta \, d\theta = \int_{1}^{-1} u^2 \, du = [\frac{u^3}{3}]_{1}^{-1} = \frac{(-1)^3}{3} - \frac{1^3}{3} = -\frac{1}{3} - \frac{1}{3} = -\frac{2}{3}$.
Thus,$I = 2 - \frac{2}{3} = \frac{4}{3}$.

(A) We evaluate the integral $I = \int_{0}^{3} \frac{3x + 1}{x^2 + 9} dx$.
Split the integral into two parts: $I = \frac{3}{2} \int_{0}^{3} \frac{2x}{x^2 + 9} dx + \int_{0}^{3} \frac{1}{x^2 + 9} dx$.
Integrating both parts,we get: $I = \left[ \frac{3}{2} \log(x^2 + 9) + \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3}$.
Applying the limits: $I = \left( \frac{3}{2} \log(9 + 9) + \frac{1}{3} \tan^{-1}(1) \right) - \left( \frac{3}{2} \log(0 + 9) + \frac{1}{3} \tan^{-1}(0) \right)$.
$I = \frac{3}{2} \log(18) + \frac{1}{3} \left(\frac{\pi}{4}\right) - \frac{3}{2} \log(9) - 0$.
$I = \frac{3}{2} \log\left(\frac{18}{9}\right) + \frac{\pi}{12} = \frac{3}{2} \log(2) + \frac{\pi}{12}$.
Since $\frac{3}{2} \log(2) = \log(2^{3/2}) = \log(2\sqrt{2})$,the final result is $\log(2\sqrt{2}) + \frac{\pi}{12}$.

(B) Let $I = \int_2^3 \frac{x + 1}{x^2(x - 1)} dx$.
Using partial fractions,we write $\frac{x + 1}{x^2(x - 1)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C}{x - 1}$.
Multiplying by $x^2(x - 1)$,we get $A(x - 1) + Bx(x - 1) + Cx^2 = x + 1$.
Setting $x = 0$,we get $A(-1) = 1 \implies A = -1$.
Setting $x = 1$,we get $C(1)^2 = 1 + 1 \implies C = 2$.
Comparing the coefficients of $x^2$,we get $B + C = 0 \implies B = -2$.
Thus,$I = \int_2^3 \left( -\frac{1}{x^2} - \frac{2}{x} + \frac{2}{x - 1} \right) dx$.
$I = \left[ \frac{1}{x} - 2\log|x| + 2\log|x - 1| \right]_2^3$.
$I = \left[ \frac{1}{x} + 2\log\left| \frac{x - 1}{x} \right| \right]_2^3$.
$I = \left( \frac{1}{3} + 2\log\left( \frac{2}{3} \right) \right) - \left( \frac{1}{2} + 2\log\left( \frac{1}{2} \right) \right)$.
$I = \left( \frac{1}{3} - \frac{1}{2} \right) + 2\log\left( \frac{2/3}{1/2} \right) = -\frac{1}{6} + 2\log\left( \frac{4}{3} \right)$.
$I = \log\left( \frac{4}{3} \right)^2 - \frac{1}{6} = \log\frac{16}{9} - \frac{1}{6}$.

(C) Given $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} dx$.
We know that $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$.
Substituting this into the integral,we get:
$I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{(\sin x + \cos x)^2}} dx$.
Since $\sin x + \cos x > 0$ for $x \in [0, \pi/2]$,$\sqrt{(\sin x + \cos x)^2} = \sin x + \cos x$.
Thus,$I = \int_0^{\pi /2} (\sin x + \cos x) dx$.
Evaluating the integral:
$I = [-\cos x + \sin x]_0^{\pi /2}$.
$I = (-\cos(\pi/2) + \sin(\pi/2)) - (-\cos(0) + \sin(0))$.
$I = (0 + 1) - (-1 + 0) = 1 + 1 = 2$.

(D) Let $I = \int_0^{\pi /8} \cos^3(4\theta) \, d\theta = \int_0^{\pi /8} \cos^2(4\theta) \cos(4\theta) \, d\theta$.
Using the identity $\cos^2(4\theta) = 1 - \sin^2(4\theta)$,we get:
$I = \int_0^{\pi /8} (1 - \sin^2(4\theta)) \cos(4\theta) \, d\theta$.
Let $t = \sin(4\theta)$. Then $dt = 4 \cos(4\theta) \, d\theta$,which implies $\cos(4\theta) \, d\theta = \frac{dt}{4}$.
When $\theta = 0$,$t = \sin(0) = 0$.
When $\theta = \frac{\pi}{8}$,$t = \sin(4 \times \frac{\pi}{8}) = \sin(\frac{\pi}{2}) = 1$.
Substituting these into the integral:
$I = \frac{1}{4} \int_0^1 (1 - t^2) \, dt = \frac{1}{4} [t - \frac{t^3}{3}]_0^1$.
$I = \frac{1}{4} (1 - \frac{1}{3}) = \frac{1}{4} (\frac{2}{3}) = \frac{1}{6}$.

(A) Let $I = \int_3^8 \frac{2 - 3x}{x\sqrt{1 + x}} \, dx$.
Substitute $1 + x = t^2$,which implies $x = t^2 - 1$ and $dx = 2t \, dt$.
When $x = 3$,$t = \sqrt{1 + 3} = 2$. When $x = 8$,$t = \sqrt{1 + 8} = 3$.
Substituting these into the integral:
$I = \int_2^3 \frac{2 - 3(t^2 - 1)}{(t^2 - 1)t} \cdot 2t \, dt = 2 \int_2^3 \frac{5 - 3t^2}{t^2 - 1} \, dt$.
$I = 2 \int_2^3 \left( \frac{2}{t^2 - 1} - 3 \right) \, dt$.
Using the formula $\int \frac{1}{t^2 - a^2} \, dt = \frac{1}{2a} \log \left| \frac{t - a}{t + a} \right|$,we get:
$I = 2 \left[ \frac{2}{2(1)} \log \left| \frac{t - 1}{t + 1} \right| - 3t \right]_2^3$.
$I = 2 \left[ \log \left| \frac{t - 1}{t + 1} \right| - 3t \right]_2^3$.
Evaluating at the limits:
$I = 2 \left[ (\log(2/4) - 9) - (\log(1/3) - 6) \right] = 2 \left[ \log(1/2) - \log(1/3) - 3 \right]$.
$I = 2 \left[ \log(3/2) - 3 \right] = 2 \log \left( \frac{3}{2e^3} \right)$.

(A) Let $I = \int_0^1 x^2 e^x dx$.
Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = x^2$ and $v = e^x$. Then $u' = 2x$ and $\int v dx = e^x$.
$I = [x^2 e^x]_0^1 - \int_0^1 2x e^x dx$.
$I = (1^2 e^1 - 0^2 e^0) - 2 \int_0^1 x e^x dx$.
$I = e - 2 \int_0^1 x e^x dx$.
Now,evaluate $\int_0^1 x e^x dx$ using integration by parts again:
$\int_0^1 x e^x dx = [x e^x]_0^1 - \int_0^1 e^x dx = (e - 0) - [e^x]_0^1 = e - (e - 1) = 1$.
Substituting this back into the expression for $I$:
$I = e - 2(1) = e - 2$.

(B) We are given ${I_1} = \int_1^2 \frac{dx}{\sqrt{1 + x^2}}$ and ${I_2} = \int_1^2 \frac{dx}{x}$.
For all $x \in (1, 2)$,we know that $1 + x^2 > x^2$.
Taking the square root on both sides,we get $\sqrt{1 + x^2} > x$.
Since both sides are positive for $x \in (1, 2)$,we can take the reciprocal,which reverses the inequality:
$\frac{1}{\sqrt{1 + x^2}} < \frac{1}{x}$.
Integrating both sides with respect to $x$ from $1$ to $2$:
$\int_1^2 \frac{dx}{\sqrt{1 + x^2}} < \int_1^2 \frac{dx}{x}$.
Therefore,${I_1} < {I_2}$,which implies ${I_2} > {I_1}$.

(B) Let $I = \int_{1/e}^{\tan x} \frac{t \, dt}{1 + t^2} + \int_{1/e}^{\cot x} \frac{dt}{t(1 + t^2)}$.
For the first integral,let $u = 1 + t^2$,then $du = 2t \, dt$,so $\int \frac{t \, dt}{1 + t^2} = \frac{1}{2} \ln(1 + t^2)$.
Evaluating from $1/e$ to $\tan x$: $\frac{1}{2} [\ln(1 + \tan^2 x) - \ln(1 + 1/e^2)] = \frac{1}{2} [\ln(\sec^2 x) - \ln(1 + 1/e^2)]$.
For the second integral,$\int \frac{dt}{t(1 + t^2)} = \int (\frac{1}{t} - \frac{t}{1 + t^2}) \, dt = \ln|t| - \frac{1}{2} \ln(1 + t^2)$.
Evaluating from $1/e$ to $\cot x$: $[\ln(\cot x) - \frac{1}{2} \ln(1 + \cot^2 x)] - [\ln(1/e) - \frac{1}{2} \ln(1 + 1/e^2)]$.
Since $1 + \cot^2 x = \csc^2 x$,this becomes $[\ln(\cot x) - \frac{1}{2} \ln(\csc^2 x)] - [\ln(1/e) - \frac{1}{2} \ln(1 + 1/e^2)]$.
Combining both parts:
$I = \frac{1}{2} \ln(\sec^2 x) - \frac{1}{2} \ln(1 + 1/e^2) + \ln(\cot x) - \frac{1}{2} \ln(\csc^2 x) - \ln(1/e) + \frac{1}{2} \ln(1 + 1/e^2)$.
$I = \ln(\sec x) + \ln(\cot x) - \ln(\csc x) - \ln(1/e)$.
$I = \ln(\frac{\sec x \cdot \cot x}{\csc x}) - \ln(1/e) = \ln(1) - \ln(1/e) = 0 - (-1) = 1$.

(A) Let $I = \int_{\pi /4}^{3\pi /4} \frac{dx}{1 + \cos x}$.
Using the identity $1 + \cos x = 2\cos^2(x/2)$,we have:
$I = \int_{\pi /4}^{3\pi /4} \frac{dx}{2\cos^2(x/2)} = \frac{1}{2} \int_{\pi /4}^{3\pi /4} \sec^2(x/2) dx$.
Integrating $\sec^2(x/2)$,we get:
$I = \frac{1}{2} [2 \tan(x/2)]_{\pi /4}^{3\pi /4} = [\tan(x/2)]_{\pi /4}^{3\pi /4}$.
Evaluating at the limits:
$I = \tan(3\pi /8) - \tan(\pi /8)$.
Using $\tan(3\pi /8) = \cot(\pi /8)$:
$I = \cot(\pi /8) - \tan(\pi /8) = \frac{\cos(\pi /8)}{\sin(\pi /8)} - \frac{\sin(\pi /8)}{\cos(\pi /8)} = \frac{\cos^2(\pi /8) - \sin^2(\pi /8)}{\sin(\pi /8)\cos(\pi /8)}$.
Using $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ and $2\sin\theta\cos\theta = \sin(2\theta)$:
$I = \frac{\cos(\pi /4)}{\frac{1}{2}\sin(\pi /4)} = 2 \cot(\pi /4) = 2(1) = 2$.

(B) We need to evaluate the definite integral: $\int_{\pi /4}^{\pi /2} \csc^2 x \, dx$.
Recall that the antiderivative of $\csc^2 x$ is $-\cot x$.
Applying the Fundamental Theorem of Calculus:
$\int_{\pi /4}^{\pi /2} \csc^2 x \, dx = [-\cot x]_{\pi /4}^{\pi /2}$
$= -(\cot(\frac{\pi}{2}) - \cot(\frac{\pi}{4}))$
Since $\cot(\frac{\pi}{2}) = 0$ and $\cot(\frac{\pi}{4}) = 1$:
$= -(0 - 1) = 1$.

(C) Let the integral be $I = \int_1^2 \frac{f'(g(x)) \cdot g'(x)}{f(g(x))} \; dx$.
Substitute $u = f(g(x))$.
Then,the differential is $du = f'(g(x)) \cdot g'(x) \; dx$.
Now,change the limits of integration:
When $x = 1$,$u = f(g(1))$.
When $x = 2$,$u = f(g(2))$.
Since $g(1) = g(2)$,it follows that $f(g(1)) = f(g(2))$.
Therefore,the integral becomes $I = \int_{f(g(1))}^{f(g(1))} \frac{1}{u} \; du$.
Since the lower and upper limits of the definite integral are identical,the value of the integral is $0$.

(C) We evaluate the definite integral $\int_{-4}^{4} |x + 2| \, dx$ by splitting the interval at $x = -2$ where the expression inside the absolute value changes sign.
Since $|x + 2| = -(x + 2)$ for $x < -2$ and $|x + 2| = (x + 2)$ for $x \ge -2$,we have:
$\int_{-4}^{4} |x + 2| \, dx = \int_{-4}^{-2} -(x + 2) \, dx + \int_{-2}^{4} (x + 2) \, dx$
Evaluating the first integral:
$\int_{-4}^{-2} -(x + 2) \, dx = -\left[ \frac{x^2}{2} + 2x \right]_{-4}^{-2} = -\left[ (\frac{4}{2} - 4) - (\frac{16}{2} - 8) \right] = -[(-2) - (0)] = 2$
Evaluating the second integral:
$\int_{-2}^{4} (x + 2) \, dx = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{4} = \left[ (\frac{16}{2} + 8) - (\frac{4}{2} - 4) \right] = [(8 + 8) - (-2)] = 16 + 2 = 18$
Adding the two results:
$2 + 18 = 20$
Thus,the correct option is $C$.

(B) We know that $1 - \cos 2x = 2 \sin^2 x$.
Substituting this into the integral,we get:
$\int_{ - \pi /2}^{\pi /2} \sqrt{\frac{1}{2}(2 \sin^2 x)} \,dx = \int_{ - \pi /2}^{\pi /2} \sqrt{\sin^2 x} \,dx = \int_{ - \pi /2}^{\pi /2} |\sin x| \,dx$.
Since $|\sin x|$ is an even function,we can write:
$2 \int_0^{\pi /2} \sin x \,dx$.
Evaluating the integral:
$2 [-\cos x]_0^{\pi /2} = 2 [-\cos(\pi/2) - (-\cos 0)] = 2 [0 + 1] = 2$.

(D) To evaluate the integral $\int_1^4 f(x) \, dx$,we split the interval $[1, 4]$ at $x = 2$ based on the definition of $f(x)$:
$\int_1^4 f(x) \, dx = \int_1^2 (4x + 3) \, dx + \int_2^4 (3x + 5) \, dx$
First part: $\int_1^2 (4x + 3) \, dx = [2x^2 + 3x]_1^2 = (2(2)^2 + 3(2)) - (2(1)^2 + 3(1)) = (8 + 6) - (2 + 3) = 14 - 5 = 9$
Second part: $\int_2^4 (3x + 5) \, dx = [\frac{3x^2}{2} + 5x]_2^4 = (\frac{3(16)}{2} + 5(4)) - (\frac{3(4)}{2} + 5(2)) = (24 + 20) - (6 + 10) = 44 - 16 = 28$
Total integral: $9 + 28 = 37$.

(A) Let $I = \int_0^1 \frac{dx}{\sqrt{1 + x^4}}$.
For $x \in [0, 1]$,we have $0 \le x^4 \le 1$.
Adding $1$ to all sides,we get $1 \le 1 + x^4 \le 2$.
Taking the square root,we get $1 \le \sqrt{1 + x^4} \le \sqrt{2}$.
Taking the reciprocal,the inequality reverses: $\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{1 + x^4}} \le 1$.
Integrating from $0$ to $1$ with respect to $x$,we get $\int_0^1 \frac{1}{\sqrt{2}} dx \le \int_0^1 \frac{dx}{\sqrt{1 + x^4}} \le \int_0^1 1 dx$.
This simplifies to $\frac{1}{\sqrt{2}}(1 - 0) \le I \le 1(1 - 0)$,which gives $\frac{1}{\sqrt{2}} \le I \le 1$.
Thus,the smallest interval is $[\frac{1}{\sqrt{2}}, 1]$.

(C) We need to evaluate the definite integral $I = \int_{-1}^{1} |1 - x| \,dx$.
Since the interval of integration is $[-1, 1]$,we observe that for all $x \in [-1, 1]$,$1 - x \ge 0$.
Specifically,when $x = 1$,$1 - x = 0$,and for $x < 1$,$1 - x > 0$.
Therefore,$|1 - x| = 1 - x$ for all $x \in [-1, 1]$.
Thus,the integral becomes:
$I = \int_{-1}^{1} (1 - x) \,dx$
$I = \left[ x - \frac{x^2}{2} \right]_{-1}^{1}$
$I = \left( 1 - \frac{1^2}{2} \right) - \left( -1 - \frac{(-1)^2}{2} \right)$
$I = \left( 1 - \frac{1}{2} \right) - \left( -1 - \frac{1}{2} \right)$
$I = \frac{1}{2} - \left( -\frac{3}{2} \right)$
$I = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$.
Hence,the correct option is $C$.

(B) The function $f(x) = |1 - x^2|$ changes its sign at $x = -1$ and $x = 1$.
We can split the integral as:
$\int_{-2}^{2} |1 - x^2| \, dx = \int_{-2}^{-1} |1 - x^2| \, dx + \int_{-1}^{1} |1 - x^2| \, dx + \int_{1}^{2} |1 - x^2| \, dx$
In the interval $[-2, -1]$,$1 - x^2 \le 0$,so $|1 - x^2| = -(1 - x^2) = x^2 - 1$.
In the interval $[-1, 1]$,$1 - x^2 \ge 0$,so $|1 - x^2| = 1 - x^2$.
In the interval $[1, 2]$,$1 - x^2 \le 0$,so $|1 - x^2| = -(1 - x^2) = x^2 - 1$.
Thus,the integral becomes:
$\int_{-2}^{-1} (x^2 - 1) \, dx + \int_{-1}^{1} (1 - x^2) \, dx + \int_{1}^{2} (x^2 - 1) \, dx$
$= [\frac{x^3}{3} - x]_{-2}^{-1} + [x - \frac{x^3}{3}]_{-1}^{1} + [\frac{x^3}{3} - x]_{1}^{2}$
$= ((\frac{-1}{3} + 1) - (\frac{-8}{3} + 2)) + ((1 - \frac{1}{3}) - (-1 + \frac{1}{3})) + ((\frac{8}{3} - 2) - (\frac{1}{3} - 1))$
$= (\frac{2}{3} - (-\frac{2}{3})) + (\frac{2}{3} - (-\frac{2}{3})) + (\frac{2}{3} - (-\frac{2}{3}))$
$= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = \frac{12}{3} = 4.$

(B) Let $I = \int_0^{\pi /2} {\left| {\sin \left( {x - \frac{\pi }{4}} \right)} \right|} \,dx$.
Since $\sin(x - \pi/4) \le 0$ for $x \in [0, \pi/4]$ and $\sin(x - \pi/4) \ge 0$ for $x \in [\pi/4, \pi/2]$,we split the integral:
$I = - \int_0^{\pi /4} {\sin \left( {x - \frac{\pi }{4}} \right)dx} + \int_{\pi /4}^{\pi /2} {\sin \left( {x - \frac{\pi }{4}} \right)dx}$.
Evaluating the integrals:
$I = - [-\cos(x - \pi/4)]_0^{\pi/4} + [-\cos(x - \pi/4)]_{\pi/4}^{\pi/2}$.
$I = [\cos(x - \pi/4)]_0^{\pi/4} - [\cos(x - \pi/4)]_{\pi/4}^{\pi/2}$.
$I = (\cos(0) - \cos(-\pi/4)) - (\cos(\pi/4) - \cos(0))$.
$I = (1 - 1/\sqrt{2}) - (1/\sqrt{2} - 1)$.
$I = 1 - 1/\sqrt{2} - 1/\sqrt{2} + 1 = 2 - 2/\sqrt{2} = 2 - \sqrt{2}$.

(C) We need to evaluate the integral $I = \int_0^\pi |\cos x| \, dx$.
Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral:
$I = \int_0^{\pi/2} \cos x \, dx + \int_{\pi/2}^\pi (-\cos x) \, dx$.
Evaluating the first part: $\int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.
Evaluating the second part: $\int_{\pi/2}^\pi -\cos x \, dx = -[\sin x]_{\pi/2}^\pi = -(\sin(\pi) - \sin(\pi/2)) = -(0 - 1) = 1$.
Adding both parts: $I = 1 + 1 = 2$.

(NONE) Let $I = \int_{-\pi/4}^{\pi/4} \sin^4 x \, dx$. Since $f(x) = \sin^4 x$ is an even function,$I = 2 \int_0^{\pi/4} \sin^4 x \, dx$.
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we have $\sin^4 x = (\frac{1 - \cos 2x}{2})^2 = \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x)$.
Further,$\cos^2 2x = \frac{1 + \cos 4x}{2}$,so $\sin^4 x = \frac{1}{4}(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}) = \frac{1}{4}(\frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x) = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.
Integrating this,$I = 2 \int_0^{\pi/4} (\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x) \, dx = 2 [\frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x]_0^{\pi/4}$.
Evaluating at the limits: $I = 2 [(\frac{3}{8} \cdot \frac{\pi}{4} - \frac{1}{4}\sin \frac{\pi}{2} + \frac{1}{32}\sin \pi) - (0)] = 2 [\frac{3\pi}{32} - \frac{1}{4}] = \frac{3\pi}{16} - \frac{1}{2}$.

(B) To evaluate the integral $I = \int_0^{1.5} [x^2] \, dx$,we break the interval $[0, 1.5]$ based on the values where $x^2$ is an integer.
$x^2 = 1 \implies x = 1$
$x^2 = 2 \implies x = \sqrt{2} \approx 1.414$
Thus,we split the integral as follows:
$I = \int_0^1 [x^2] \, dx + \int_1^{\sqrt{2}} [x^2] \, dx + \int_{\sqrt{2}}^{1.5} [x^2] \, dx$
For $0 \le x < 1$,$0 \le x^2 < 1$,so $[x^2] = 0$.
For $1 \le x < \sqrt{2}$,$1 \le x^2 < 2$,so $[x^2] = 1$.
For $\sqrt{2} \le x < 1.5$,$2 \le x^2 < 2.25$,so $[x^2] = 2$.
Substituting these values:
$I = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{1.5} 2 \, dx$
$I = 0 + [x]_1^{\sqrt{2}} + 2[x]_{\sqrt{2}}^{1.5}$
$I = (\sqrt{2} - 1) + 2(1.5 - \sqrt{2})$
$I = \sqrt{2} - 1 + 3 - 2\sqrt{2}$
$I = 2 - \sqrt{2}$

(B) We need to evaluate the integral $I = \int_{1/e}^e |\log x| \, dx$.
Since $\log x < 0$ for $x \in [1/e, 1)$ and $\log x \ge 0$ for $x \in [1, e]$,we split the integral:
$I = \int_{1/e}^1 -\log x \, dx + \int_1^e \log x \, dx$.
Using the integral formula $\int \log x \, dx = x \log x - x + C$:
$I = -[x \log x - x]_{1/e}^1 + [x \log x - x]_1^e$.
Evaluating the first part: $-[(1 \cdot 0 - 1) - (\frac{1}{e} \log(1/e) - \frac{1}{e})] = -[-1 - (-\frac{1}{e} - \frac{1}{e})] = -[-1 + \frac{2}{e}] = 1 - \frac{2}{e}$.
Evaluating the second part: $[(e \log e - e) - (1 \log 1 - 1)] = [(e - e) - (0 - 1)] = 0 - (-1) = 1$.
Adding both parts: $I = (1 - \frac{2}{e}) + 1 = 2 - \frac{2}{e} = 2 \left( 1 - \frac{1}{e} \right)$.

(D) We are given the integral $I = \int_{0}^{\pi /2} \{ x - [\sin x] \} \,dx$.
Since $\{ x \} = x - [x]$,we can rewrite the expression as $\{ x - [\sin x] \} = x - [\sin x] - [x - [\sin x]]$.
However,for $x \in [0, \pi/2]$,we know that $0 \le \sin x < 1$,so $[\sin x] = 0$.
Thus,the integral simplifies to $I = \int_{0}^{\pi /2} \{ x - 0 \} \,dx = \int_{0}^{\pi /2} \{ x \} \,dx$.
Since $0 \le x \le \pi/2$ and $\pi/2 \approx 1.57$,the interval $[0, \pi/2]$ contains the integer $1$.
We split the integral: $I = \int_{0}^{1} x \,dx + \int_{1}^{\pi /2} (x - 1) \,dx$.
Calculating the first part: $\int_{0}^{1} x \,dx = [\frac{x^2}{2}]_{0}^{1} = \frac{1}{2}$.
Calculating the second part: $\int_{1}^{\pi /2} (x - 1) \,dx = [\frac{x^2}{2} - x]_{1}^{\pi /2} = (\frac{\pi^2}{8} - \frac{\pi}{2}) - (\frac{1}{2} - 1) = \frac{\pi^2}{8} - \frac{\pi}{2} + \frac{1}{2}$.
Adding them together: $I = \frac{1}{2} + \frac{\pi^2}{8} - \frac{\pi}{2} + \frac{1}{2} = \frac{\pi^2}{8} - \frac{\pi}{2} + 1$.
Since this result is not among the options,the correct choice is $D$.

(C) Let $I = \int_{\pi}^{2\pi} [2\sin x] \, dx$. In the interval $[\pi, 2\pi]$,$\sin x$ ranges from $0$ to $-1$ and back to $0$. Thus,$2\sin x$ ranges from $0$ to $-2$ and back to $0$.
We split the integral based on the values of $[2\sin x]$:
$1$. For $x \in [\pi, 7\pi/6]$,$2\sin x \in [-1, 0]$,so $[2\sin x] = -1$.
$2$. For $x \in [7\pi/6, 3\pi/2]$,$2\sin x \in [-2, -1]$,so $[2\sin x] = -2$.
$3$. For $x \in [3\pi/2, 11\pi/6]$,$2\sin x \in [-2, -1]$,so $[2\sin x] = -2$.
$4$. For $x \in [11\pi/6, 2\pi]$,$2\sin x \in [-1, 0]$,so $[2\sin x] = -1$.
$I = \int_{\pi}^{7\pi/6} (-1) \, dx + \int_{7\pi/6}^{3\pi/2} (-2) \, dx + \int_{3\pi/2}^{11\pi/6} (-2) \, dx + \int_{11\pi/6}^{2\pi} (-1) \, dx$
$I = -1(\frac{7\pi}{6} - \pi) - 2(\frac{3\pi}{2} - \frac{7\pi}{6}) - 2(\frac{11\pi}{6} - \frac{3\pi}{2}) - 1(2\pi - \frac{11\pi}{6})$
$I = -(\frac{\pi}{6}) - 2(\frac{2\pi}{6}) - 2(\frac{2\pi}{6}) - (\frac{\pi}{6})$
$I = -\frac{\pi}{6} - \frac{4\pi}{6} - \frac{4\pi}{6} - \frac{\pi}{6} = -\frac{10\pi}{6} = -\frac{5\pi}{3}$.

(D) Let $I = \int_1^5 (|x - 3| + |1 - x|) \, dx$.
Since $x \in [1, 5]$,$|1 - x| = x - 1$.
So,$I = \int_1^5 |x - 3| \, dx + \int_1^5 (x - 1) \, dx$.
For the first part,$\int_1^5 |x - 3| \, dx = \int_1^3 -(x - 3) \, dx + \int_3^5 (x - 3) \, dx$.
$= [-\frac{(x - 3)^2}{2}]_1^3 + [\frac{(x - 3)^2}{2}]_3^5 = (0 - (-2)) + (2 - 0) = 2 + 2 = 4$.
For the second part,$\int_1^5 (x - 1) \, dx = [\frac{x^2}{2} - x]_1^5 = (\frac{25}{2} - 5) - (\frac{1}{2} - 1) = 7.5 - (-0.5) = 8$.
Therefore,$I = 4 + 8 = 12$.

(A) Let $I = \int_0^\pi \cos mx \sin nx \,dx$.
Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have $\cos mx \sin nx = \frac{1}{2} [\sin(n+m)x - \sin(m-n)x] = \frac{1}{2} [\sin(n+m)x + \sin(n-m)x]$.
Integrating this,we get:
$I = \frac{1}{2} \int_0^\pi [\sin(n+m)x + \sin(n-m)x] \,dx$
$I = \frac{1}{2} \left[ -\frac{\cos(n+m)x}{n+m} - \frac{\cos(n-m)x}{n-m} \right]_0^\pi$
$I = -\frac{1}{2} \left[ \left( \frac{\cos(n+m)\pi}{n+m} + \frac{\cos(n-m)\pi}{n-m} \right) - \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right]$
Since $(n-m)$ is odd,$(n+m)$ is also odd. Thus,$\cos(n+m)\pi = -1$ and $\cos(n-m)\pi = -1$.
$I = -\frac{1}{2} \left[ \left( \frac{-1}{n+m} - \frac{1}{n-m} \right) - \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right]$
$I = -\frac{1}{2} \left[ -2 \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right] = \frac{1}{n+m} + \frac{1}{n-m}$
$I = \frac{(n-m) + (n+m)}{n^2 - m^2} = \frac{2n}{n^2 - m^2}$.

(B) Given $I = \int_0^{100\pi} \sqrt{1 - \cos 2x} \, dx$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,we have $\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$.
So,$I = \sqrt{2} \int_0^{100\pi} |\sin x| \, dx$.
Since $|\sin x|$ is a periodic function with period $\pi$,we can write $\int_0^{100\pi} |\sin x| \, dx = 100 \int_0^{\pi} |\sin x| \, dx$.
In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$.
Thus,$I = 100\sqrt{2} \int_0^{\pi} \sin x \, dx$.
Evaluating the integral: $\int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$.
Therefore,$I = 100\sqrt{2} \times 2 = 200\sqrt{2}$.

(B) The integral is given by $I = \int_0^2 x^2 [x] \, dx$.
Since $[x]$ is the greatest integer function,we split the interval $[0, 2]$ into $[0, 1)$ and $[1, 2]$.
For $0 \le x < 1$,$[x] = 0$.
For $1 \le x < 2$,$[x] = 1$.
Thus,$I = \int_0^1 x^2(0) \, dx + \int_1^2 x^2(1) \, dx$.
$I = 0 + \int_1^2 x^2 \, dx$.
$I = \left[ \frac{x^3}{3} \right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$.

(D) We need to evaluate the integral $I = \int_{0}^{2\pi} |\sin x| \, dx$.
Since $|\sin x| = \sin x$ for $x \in [0, \pi]$ and $|\sin x| = -\sin x$ for $x \in [\pi, 2\pi]$,we split the integral:
$I = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx$
$I = [-\cos x]_{0}^{\pi} + [\cos x]_{\pi}^{2\pi}$
$I = (-(\cos \pi - \cos 0)) + (\cos 2\pi - \cos \pi)$
$I = (-(-1 - 1)) + (1 - (-1))$
$I = (2) + (2) = 4$.

(C) The function $f(\theta) = |\sin^3 \theta|$ is periodic with period $\pi$.
Thus,$\int_0^{2\pi} |\sin^3 \theta| \, d\theta = 2 \int_0^{\pi} |\sin^3 \theta| \, d\theta$.
Since $\sin \theta \ge 0$ for $\theta \in [0, \pi]$,we have $|\sin^3 \theta| = \sin^3 \theta$.
Using the identity $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$,we get $\sin^3 \theta = \frac{3\sin \theta - \sin 3\theta}{4}$.
Therefore,the integral becomes $2 \int_0^{\pi} \frac{3\sin \theta - \sin 3\theta}{4} \, d\theta = \frac{1}{2} \int_0^{\pi} (3\sin \theta - \sin 3\theta) \, d\theta$.
Evaluating the integral: $\frac{1}{2} [-3\cos \theta + \frac{1}{3}\cos 3\theta]_0^{\pi}$.
$= \frac{1}{2} [(-3\cos \pi + \frac{1}{3}\cos 3\pi) - (-3\cos 0 + \frac{1}{3}\cos 0)]$.
$= \frac{1}{2} [(3 - 1/3) - (-3 + 1/3)] = \frac{1}{2} [8/3 - (-8/3)] = \frac{1}{2} [16/3] = 8/3$.

(A) We need to evaluate the definite integral $I = \int_{-1}^{2} {|x|\,dx}$.
Since the integrand $|x|$ changes its definition at $x = 0$,we split the integral into two parts:
$I = \int_{-1}^{0} {|x|\,dx} + \int_{0}^{2} {|x|\,dx}$.
For $x \in [-1, 0]$,$|x| = -x$,and for $x \in [0, 2]$,$|x| = x$.
So,$I = \int_{-1}^{0} {-x\,dx} + \int_{0}^{2} {x\,dx}$.
Evaluating the first part: $\int_{-1}^{0} {-x\,dx} = -\left[ \frac{x^2}{2} \right]_{-1}^{0} = -\left( 0 - \frac{(-1)^2}{2} \right) = -\left( 0 - \frac{1}{2} \right) = \frac{1}{2}$.
Evaluating the second part: $\int_{0}^{2} {x\,dx} = \left[ \frac{x^2}{2} \right]_{0}^{2} = \left( \frac{2^2}{2} - 0 \right) = \frac{4}{2} = 2$.
Adding both parts: $I = \frac{1}{2} + 2 = \frac{5}{2}$.

(B) Let $I = \int_{0}^{3} {|2 - x| \, dx}$.
Since $|2 - x| = 2 - x$ for $x \le 2$ and $|2 - x| = -(2 - x) = x - 2$ for $x > 2$,we split the integral at $x = 2$:
$I = \int_{0}^{2} {(2 - x) \, dx} + \int_{2}^{3} {(x - 2) \, dx}$
Evaluating the first integral:
$\int_{0}^{2} {(2 - x) \, dx} = [2x - \frac{x^2}{2}]_{0}^{2} = (4 - 2) - (0 - 0) = 2$
Evaluating the second integral:
$\int_{2}^{3} {(x - 2) \, dx} = [\frac{x^2}{2} - 2x]_{2}^{3} = (\frac{9}{2} - 6) - (\frac{4}{2} - 4) = (\frac{9-12}{2}) - (2 - 4) = -\frac{3}{2} - (-2) = -\frac{3}{2} + 2 = \frac{1}{2}$
Adding the two parts:
$I = 2 + \frac{1}{2} = \frac{5}{2}$.

(B) We need to evaluate the integral $I = \int_{0}^{1} |3x^2 - 1| dx$.
First,we find the critical point where $3x^2 - 1 = 0$,which gives $x^2 = 1/3$,so $x = 1/\sqrt{3}$ (since $x \in [0, 1]$).
For $0 \le x < 1/\sqrt{3}$,$3x^2 - 1 < 0$,so $|3x^2 - 1| = 1 - 3x^2$.
For $1/\sqrt{3} < x \le 1$,$3x^2 - 1 > 0$,so $|3x^2 - 1| = 3x^2 - 1$.
Thus,$I = \int_{0}^{1/\sqrt{3}} (1 - 3x^2) dx + \int_{1/\sqrt{3}}^{1} (3x^2 - 1) dx$.
Evaluating the integrals:
$I = [x - x^3]_{0}^{1/\sqrt{3}} + [x^3 - x]_{1/\sqrt{3}}^{1}$.
$I = (\frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}}) - (0) + (1 - 1) - (\frac{1}{3\sqrt{3}} - \frac{1}{\sqrt{3}})$.
$I = \frac{2}{3\sqrt{3}} - (\frac{1-3}{3\sqrt{3}}) = \frac{2}{3\sqrt{3}} + \frac{2}{3\sqrt{3}} = \frac{4}{3\sqrt{3}}$.

(B) Let $I = \int_{1}^{5} [|x - 3|] \, dx$.
We split the integral based on the definition of the absolute value and the greatest integer function:
$I = \int_{1}^{3} [-(x - 3)] \, dx + \int_{3}^{5} [x - 3] \, dx$.
For the first part,$\int_{1}^{3} [3 - x] \, dx$:
When $1 \le x < 2$,$1 < 3 - x \le 2$,so $[3 - x] = 1$.
When $2 \le x < 3$,$0 < 3 - x \le 1$,so $[3 - x] = 0$.
Thus,$\int_{1}^{3} [3 - x] \, dx = \int_{1}^{2} 1 \, dx + \int_{2}^{3} 0 \, dx = [x]_{1}^{2} = 2 - 1 = 1$.
For the second part,$\int_{3}^{5} [x - 3] \, dx$:
When $3 \le x < 4$,$0 \le x - 3 < 1$,so $[x - 3] = 0$.
When $4 \le x < 5$,$1 \le x - 3 < 2$,so $[x - 3] = 1$.
Thus,$\int_{3}^{5} [x - 3] \, dx = \int_{3}^{4} 0 \, dx + \int_{4}^{5} 1 \, dx = [x]_{4}^{5} = 5 - 4 = 1$.
Therefore,$I = 1 + 1 = 2$.

(D) Let $I = \int_{-\pi}^{\pi} (\cos ax - \sin bx)^2 dx$.
Expanding the integrand,we get $I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx - 2\cos ax \sin bx) dx$.
Using the linearity of the integral,$I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx) dx - 2\int_{-\pi}^{\pi} \cos ax \sin bx dx$.
Since $\cos ax \sin bx$ is an odd function,the integral $\int_{-\pi}^{\pi} \cos ax \sin bx dx = 0$.
Thus,$I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx) dx$.
Since the integrand is an even function,$I = 2 \int_{0}^{\pi} (\cos^2 ax + \sin^2 bx) dx$.
Using the identities $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ and $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we have:
$I = 2 \int_{0}^{\pi} (\frac{1 + \cos 2ax}{2} + \frac{1 - \cos 2bx}{2}) dx = \int_{0}^{\pi} (2 + \cos 2ax - \cos 2bx) dx$.
Integrating term by term,$I = [2x + \frac{\sin 2ax}{2a} - \frac{\sin 2bx}{2b}]_{0}^{\pi}$.
Since $a$ and $b$ are integers,$\sin(2a\pi) = 0$ and $\sin(2b\pi) = 0$.
Therefore,$I = 2\pi + 0 - 0 = 2\pi$.