Fundamental definite integration Questions in English

(B) Let $I = \int_{0}^{\infty} \frac{x}{(1+x)(x^{2}+1)} dx$.
Using partial fractions,we write $\frac{x}{(1+x)(x^{2}+1)} = \frac{A}{1+x} + \frac{Bx+C}{x^{2}+1}$.
Equating numerators: $x = A(x^{2}+1) + (Bx+C)(1+x) = (A+B)x^{2} + (B+C)x + (A+C)$.
Comparing coefficients: $A+B=0$,$B+C=1$,$A+C=0$.
Solving these,we get $A = -\frac{1}{2}$,$B = \frac{1}{2}$,and $C = \frac{1}{2}$.
Thus,$I = \int_{0}^{\infty} \left( -\frac{1}{2(1+x)} + \frac{x+1}{2(x^{2}+1)} \right) dx = -\frac{1}{2} \int_{0}^{\infty} \frac{dx}{1+x} + \frac{1}{4} \int_{0}^{\infty} \frac{2x}{x^{2}+1} dx + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x^{2}+1}$.
Evaluating the integrals: $I = \left[ -\frac{1}{2} \ln(1+x) + \frac{1}{4} \ln(x^{2}+1) + \frac{1}{2} \tan^{-1}(x) \right]_{0}^{\infty}$.
Combining the logarithmic terms: $I = \left[ \frac{1}{4} \ln \left( \frac{(x^{2}+1)}{(1+x)^{2}} \right) + \frac{1}{2} \tan^{-1}(x) \right]_{0}^{\infty}$.
As $x \to \infty$,$\frac{x^{2}+1}{(1+x)^{2}} \to 1$,so $\ln(1) = 0$.
At $x=0$,$\frac{1}{4} \ln(1) + \frac{1}{2} \tan^{-1}(0) = 0$.
Therefore,$I = 0 + \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}$.

(B) Let $I = \int_{5}^{10} \frac{1}{(x-1)(x-2)} dx$.
Using partial fractions,we write $\frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.
Solving for constants,we get $A = -1$ and $B = 1$.
Thus,$I = \int_{5}^{10} \left( \frac{-1}{x-1} + \frac{1}{x-2} \right) dx$.
Integrating,we get $I = [-\log|x-1| + \log|x-2|]_{5}^{10}$.
$I = [\log|\frac{x-2}{x-1}|]_{5}^{10}$.
Substituting the limits,$I = \log|\frac{10-2}{10-1}| - \log|\frac{5-2}{5-1}|$.
$I = \log(\frac{8}{9}) - \log(\frac{3}{4})$.
$I = \log(\frac{8/9}{3/4}) = \log(\frac{8}{9} \times \frac{4}{3}) = \log(\frac{32}{27})$.

(B) To evaluate the integral $\int_0^1 x e^x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x$ and $dv = e^x \, dx$.
Then $du = dx$ and $v = e^x$.
Applying the formula:
$\int_0^1 x e^x \, dx = [x e^x]_0^1 - \int_0^1 e^x \, dx$
$= (1 \cdot e^1 - 0 \cdot e^0) - [e^x]_0^1$
$= (e - 0) - (e^1 - e^0)$
$= e - (e - 1)$
$= e - e + 1$
$= 1$.
Thus,the correct option is $B$.

(C) We have the integral $I = \int_0^1 (0.001)^{\frac{x}{3}} e^x \, dx$.
First,simplify the term $(0.001)^{\frac{x}{3}}$.
Since $0.001 = 10^{-3}$,we have $(10^{-3})^{\frac{x}{3}} = 10^{-x} = \frac{1}{10^x}$.
Thus,the integral becomes $I = \int_0^1 \frac{e^x}{10^x} \, dx = \int_0^1 \left(\frac{e}{10}\right)^x \, dx$.
Using the formula $\int a^x \, dx = \frac{a^x}{\ln a} + C$,we get:
$I = \left[ \frac{(\frac{e}{10})^x}{\ln(\frac{e}{10})} \right]_0^1 = \frac{\frac{e}{10} - 1}{\ln e - \ln 10} = \frac{\frac{e-10}{10}}{1 - \ln 10}$.
Therefore,$I = \frac{e-10}{10(1-\ln 10)}$.
This matches option $C$.

(B) Since $f(x) = \sin^2 x$ is an even function,we can write:
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx$
Using the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$:
$2 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos(2x)}{2} \, dx = \int_{0}^{\frac{\pi}{4}} (1 - \cos(2x)) \, dx$
Integrating term by term:
$[x - \frac{\sin(2x)}{2}]_{0}^{\frac{\pi}{4}}$
Substituting the limits:
$(\frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2}) - (0 - \frac{\sin(0)}{2})$
$= \frac{\pi}{4} - \frac{1}{2} - 0 = \frac{\pi}{4} - \frac{1}{2}$
Thus,the correct option is $B$.

(C) We are given the integral $ I = \int_{1}^{3} \left(\frac{x^{2}+1}{4x}\right)^{-1} dx $.
Simplifying the integrand,we get $ I = \int_{1}^{3} \frac{4x}{x^{2}+1} dx $.
Let $ u = x^{2}+1 $,then $ du = 2x dx $,which implies $ 2 du = 4x dx $.
When $ x = 1 $,$ u = 1^{2}+1 = 2 $.
When $ x = 3 $,$ u = 3^{2}+1 = 10 $.
Substituting these into the integral,we get $ I = \int_{2}^{10} \frac{2}{u} du $.
$ I = 2 [\ln |u|]_{2}^{10} $.
$ I = 2 (\ln 10 - \ln 2) $.
$ I = 2 \ln \left(\frac{10}{2}\right) = 2 \ln 5 $.
Using the property $ n \ln a = \ln(a^{n}) $,we get $ I = \ln(5^{2}) = \ln 25 $.
Thus,the correct option is $ C $.

(C) To evaluate the integral $I = \int_0^1 x(1-x)^n dx$,we use the property of definite integrals: $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.
Applying this property,we get:
$I = \int_0^1 (1-x)(1-(1-x))^n dx$
$I = \int_0^1 (1-x)x^n dx$
$I = \int_0^1 (x^n - x^{n+1}) dx$
Now,integrate term by term:
$I = [\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}]_0^1$
$I = \frac{1}{n+1} - \frac{1}{n+2}$
$I = \frac{(n+2) - (n+1)}{(n+1)(n+2)}$
$I = \frac{1}{n^2 + 3n + 2}$
Thus,the correct option is $C$.

(A) We need to evaluate the definite integral $I = \int_{0}^{\pi} \sin x \, dx$.
The antiderivative of $\sin x$ is $-\cos x$.
Applying the Fundamental Theorem of Calculus:
$I = [-\cos x]_{0}^{\pi}$
$I = -(\cos \pi - \cos 0)$
$I = -(-1 - 1)$
$I = -(-2)$
$I = 2$
Therefore,the correct option is $A$.

(B) To evaluate the integral $I = \int_0^1 \sin^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \sin^{-1} x$ and $dv = dx$.
Then $du = \frac{1}{\sqrt{1 - x^2}} \, dx$ and $v = x$.
Applying the formula:
$I = [x \sin^{-1} x]_0^1 - \int_0^1 \frac{x}{\sqrt{1 - x^2}} \, dx$.
Evaluating the first part: $[1 \cdot \sin^{-1}(1) - 0 \cdot \sin^{-1}(0)] = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}$.
For the second part,let $t = 1 - x^2$,so $dt = -2x \, dx$ or $x \, dx = -\frac{1}{2} dt$.
When $x=0, t=1$; when $x=1, t=0$.
$-\int_1^0 \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int_0^1 t^{-1/2} \, dt = -\frac{1}{2} [2\sqrt{t}]_0^1 = -\frac{1}{2} [2] = -1$.
Thus,$I = \frac{\pi}{2} - 1$.

(B) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x $.
Divide numerator and denominator by $ \cos^{2} x $:
$ I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^{2} x}{a^{2} \tan^{2} x + b^{2}} d x $.
Let $ \tan x = t $,then $ \sec^{2} x d x = d t $.
When $ x = 0, t = 0 $ and when $ x = \frac{\pi}{2}, t \to \infty $.
$ I = \int_{0}^{\infty} \frac{d t}{a^{2} t^{2} + b^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{d t}{t^{2} + (b/a)^{2}} $.
Using the formula $ \int \frac{d x}{x^{2} + k^{2}} = \frac{1}{k} \tan^{-1}(\frac{x}{k}) $:
$ I = \frac{1}{a^{2}} \cdot \frac{1}{(b/a)} \left[ \tan^{-1} \left( \frac{t}{b/a} \right) \right]_{0}^{\infty} $.
$ I = \frac{1}{ab} \left( \tan^{-1}(\infty) - \tan^{-1}(0) \right) = \frac{1}{ab} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{2ab} $.

(A) Given the integral $ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{1+\cos 2x} $.
Using the trigonometric identity $ 1+\cos 2x = 2\cos^2 x $,we have:
$ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{2\cos^2 x} = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \, dx $.
Since $ f(x) = \sec^2 x $ is an even function,we can use the property $ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx $.
Thus,$ I = \frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx = \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx $.
Evaluating the integral,we get $ [\tan x]_{0}^{\frac{\pi}{4}} = \tan(\frac{\pi}{4}) - \tan(0) = 1 - 0 = 1 $.

(A) We need to evaluate the integral $ I = \int_{-5}^{5} |x+2| \, dx $.
Since $|x+2| = -(x+2)$ for $x < -2$ and $|x+2| = (x+2)$ for $x \ge -2$,we split the integral at $x = -2$:
$ I = \int_{-5}^{-2} -(x+2) \, dx + \int_{-2}^{5} (x+2) \, dx $
Evaluating the first part:
$ -\left[ \frac{x^2}{2} + 2x \right]_{-5}^{-2} = -\left( (\frac{4}{2} - 4) - (\frac{25}{2} - 10) \right) = -\left( -2 - 2.5 \right) = 4.5 $
Evaluating the second part:
$ \left[ \frac{x^2}{2} + 2x \right]_{-2}^{5} = \left( (\frac{25}{2} + 10) - (\frac{4}{2} - 4) \right) = (12.5 + 10) - (-2) = 22.5 + 2 = 24.5 $
Adding both parts:
$ I = 4.5 + 24.5 = 29 $

(D) We are given $I_{1} = \int_{0}^{\pi / 2} x \sin x \, dx$ and $I_{2} = \int_{0}^{\pi / 2} x \cos x \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
For $I_{1}$,let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$.
$I_{1} = [x(-\cos x)]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} (-\cos x) \, dx = [0 - 0] + [\sin x]_{0}^{\pi / 2} = 1 - 0 = 1$.
For $I_{2}$,let $u = x$ and $dv = \cos x \, dx$. Then $du = dx$ and $v = \sin x$.
$I_{2} = [x \sin x]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} \sin x \, dx = [\frac{\pi}{2} \sin(\frac{\pi}{2}) - 0] - [-\cos x]_{0}^{\pi / 2} = \frac{\pi}{2} - [-(0 - 1)] = \frac{\pi}{2} - 1$.
Adding the two results,$I_{1} + I_{2} = 1 + (\frac{\pi}{2} - 1) = \frac{\pi}{2}$.
Thus,the correct option is $D$.

(C) Let $I = \int_0^{\pi / 2} \sqrt{\sin \theta} \cos^3 \theta d \theta$.
Substitute $\sin \theta = t$,then $\cos \theta d \theta = dt$.
When $\theta = 0$,$t = 0$. When $\theta = \frac{\pi}{2}$,$t = 1$.
We can rewrite $\cos^3 \theta d \theta$ as $\cos^2 \theta (\cos \theta d \theta) = (1 - \sin^2 \theta) \cos \theta d \theta = (1 - t^2) dt$.
Thus,$I = \int_0^1 \sqrt{t} (1 - t^2) dt = \int_0^1 (t^{1/2} - t^{5/2}) dt$.
Integrating term by term: $I = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} \right]_0^1$.
$I = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} \right]_0^1$.
$I = \left( \frac{2}{3} - \frac{2}{7} \right) - (0 - 0) = \frac{14 - 6}{21} = \frac{8}{21}$.

(D) Let $I = \int_0^{\pi / 2} \frac{\cos x \sin x}{1+\sin x} d x$.
We can rewrite the numerator as $\cos x(1+\sin x - 1)$.
Then,$I = \int_0^{\pi / 2} \frac{\cos x(1+\sin x) - \cos x}{1+\sin x} d x$.
This simplifies to $I = \int_0^{\pi / 2} \cos x d x - \int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x$.
Evaluating the first integral: $\int_0^{\pi / 2} \cos x d x = [\sin x]_0^{\pi / 2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.
For the second integral,let $t = 1 + \sin x$,then $dt = \cos x dx$.
When $x = 0, t = 1$; when $x = \pi/2, t = 2$.
So,$\int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x = \int_1^2 \frac{dt}{t} = [\log |t|]_1^2 = \log 2 - \log 1 = \log 2$.
Therefore,$I = 1 - \log 2$.

(A) Let $ I = \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx $.
Rationalizing the integrand:
$ I = \int_{0}^{1} \sqrt{\frac{(1+x)^2}{(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 $ 1-x^2 = t $,then $ -2x \, dx = dt $,so $ x \, dx = -\frac{1}{2} \, dt $.
When $ x=0, t=1 $ and when $ x=1, t=0 $.
$ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx = \int_{1}^{0} \frac{-1/2}{\sqrt{t}} \, dt = \frac{1}{2} \int_{0}^{1} t^{-1/2} \, dt = \frac{1}{2} [2\sqrt{t}]_{0}^{1} = \frac{1}{2} (2) = 1 $.
Therefore,$ I = \frac{\pi}{2} + 1 $.

(A) Let $ I = \int_{0}^{1/2} \frac{dx}{(1+x^{2}) \sqrt{1-x^{2}}} $.
Substitute $ x = \sin \theta $,then $ dx = \cos \theta d\theta $.
When $ x = 0, \theta = 0 $. When $ x = 1/2, \theta = \pi/6 $.
The integral becomes $ I = \int_{0}^{\pi/6} \frac{\cos \theta d\theta}{(1+\sin^{2} \theta) \sqrt{1-\sin^{2} \theta}} = \int_{0}^{\pi/6} \frac{\cos \theta d\theta}{(1+\sin^{2} \theta) \cos \theta} = \int_{0}^{\pi/6} \frac{d\theta}{1+\sin^{2} \theta} $.
Divide numerator and denominator by $ \cos^{2} \theta $: $ I = \int_{0}^{\pi/6} \frac{\sec^{2} \theta d\theta}{\sec^{2} \theta + \tan^{2} \theta} = \int_{0}^{\pi/6} \frac{\sec^{2} \theta d\theta}{1 + 2\tan^{2} \theta} $.
Let $ u = \sqrt{2} \tan \theta $,then $ du = \sqrt{2} \sec^{2} \theta d\theta $,so $ \sec^{2} \theta d\theta = \frac{du}{\sqrt{2}} $.
Limits: $ \theta = 0 \implies u = 0 $; $ \theta = \pi/6 \implies u = \sqrt{2} \tan(\pi/6) = \sqrt{2} \cdot \frac{1}{\sqrt{3}} = \sqrt{\frac{2}{3}} $.
Thus,$ I = \int_{0}^{\sqrt{2/3}} \frac{1}{1+u^{2}} \cdot \frac{du}{\sqrt{2}} = \frac{1}{\sqrt{2}} [\tan^{-1} u]_{0}^{\sqrt{2/3}} = \frac{1}{\sqrt{2}} \tan^{-1} \sqrt{\frac{2}{3}} $.

(A) Let $I = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$.
So,the integral becomes $I = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3 + 1 - (\sin x - \cos x)^2} d x = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{4 - (\sin x - \cos x)^2} d x$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi/4$,$t = \sin(\pi/4) - \cos(\pi/4) = 0$.
Thus,$I = \int_{-1}^{0} \frac{dt}{4 - t^2} = \int_{-1}^{0} \frac{dt}{2^2 - t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$,we get:
$I = \left[ \frac{1}{2(2)} \log \left| \frac{2+t}{2-t} \right| \right]_{-1}^{0} = \frac{1}{4} \left[ \log \left| \frac{2+0}{2-0} \right| - \log \left| \frac{2-1}{2-(-1)} \right| \right]$.
$I = \frac{1}{4} [ \log(1) - \log(1/3) ] = \frac{1}{4} [ 0 - (-\log 3) ] = \frac{1}{4} \log 3$.

(A) We know that $1 + \sin \theta = \cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2} + 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} = (\cos \frac{\theta}{2} + \sin \frac{\theta}{2})^2$.
Substituting $\theta = \frac{x}{2}$,we get $\sqrt{1 + \sin \frac{x}{2}} = |\cos \frac{x}{4} + \sin \frac{x}{4}|$.
Since $x \in [0, 2\pi]$,$\frac{x}{4} \in [0, \frac{\pi}{2}]$,where both $\sin \frac{x}{4}$ and $\cos \frac{x}{4}$ are positive.
Thus,the integral becomes $\int_0^{2\pi} (\cos \frac{x}{4} + \sin \frac{x}{4}) dx$.
Evaluating the integral: $[4 \sin \frac{x}{4} - 4 \cos \frac{x}{4}]_0^{2\pi}$.
$= (4 \sin \frac{\pi}{2} - 4 \cos \frac{\pi}{2}) - (4 \sin 0 - 4 \cos 0)$.
$= (4(1) - 4(0)) - (4(0) - 4(1)) = 4 - (-4) = 8$.

(A) We need to evaluate the integral $I = \int_1^5 (|x-3| + |1-x|) \, dx$.
Since $x \in [1, 5]$,we have $|1-x| = x-1$.
Thus,$I = \int_1^5 |x-3| \, dx + \int_1^5 (x-1) \, dx$.
For the first part,$|x-3| = 3-x$ for $x \in [1, 3]$ and $|x-3| = x-3$ for $x \in [3, 5]$.
$\int_1^3 (3-x) \, dx = [3x - \frac{x^2}{2}]_1^3 = (9 - 4.5) - (3 - 0.5) = 4.5 - 2.5 = 2$.
$\int_3^5 (x-3) \, dx = [\frac{x^2}{2} - 3x]_3^5 = (12.5 - 15) - (4.5 - 9) = -2.5 - (-4.5) = 2$.
For the second part,$\int_1^5 (x-1) \, dx = [\frac{x^2}{2} - x]_1^5 = (12.5 - 5) - (0.5 - 1) = 7.5 - (-0.5) = 8$.
Adding these results: $I = 2 + 2 + 8 = 12$.

(A) Let $I = \int_0^8 [x] dx$.
Since $[x]$ is the greatest integer function,it takes constant integer values on intervals $[n, n+1)$.
We can split the integral as:
$I = \int_0^1 0 dx + \int_1^2 1 dx + \int_2^3 2 dx + \int_3^4 3 dx + \int_4^5 4 dx + \int_5^6 5 dx + \int_6^7 6 dx + \int_7^8 7 dx$.
Evaluating each integral:
$I = 0 + 1(2-1) + 2(3-2) + 3(4-3) + 4(5-4) + 5(6-5) + 6(7-6) + 7(8-7)$.
$I = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7$.
$I = \frac{7(7+1)}{2} = \frac{7 \times 8}{2} = 28$.

(B) To evaluate the integral $ I = \int_{0}^{1} \frac{dx}{e^{x}+e^{-x}} $,we first simplify the integrand by multiplying the numerator and denominator by $ e^{x} $:
$ I = \int_{0}^{1} \frac{e^{x}}{e^{2x}+1} dx $
Now,let $ u = e^{x} $. Then $ du = e^{x} dx $.
When $ x = 0 $,$ u = e^{0} = 1 $.
When $ x = 1 $,$ u = e^{1} = e $.
Substituting these into the integral,we get:
$ I = \int_{1}^{e} \frac{du}{u^{2}+1} $
The integral of $ \frac{1}{u^{2}+1} $ is $ \tan^{-1}(u) $.
Evaluating the definite integral:
$ I = [\tan^{-1}(u)]_{1}^{e} $
$ I = \tan^{-1}(e) - \tan^{-1}(1) $
Since $ \tan^{-1}(1) = \frac{\pi}{4} $,we have:
$ I = \tan^{-1}(e) - \frac{\pi}{4} $

(B) Let $I = \int_{-1}^{2} \frac{|x|}{x} d x$.
We know that $|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$.
Splitting the integral at $x = 0$,we get:
$I = \int_{-1}^{0} \frac{-x}{x} d x + \int_{0}^{2} \frac{x}{x} d x$
$I = \int_{-1}^{0} (-1) d x + \int_{0}^{2} (1) d x$
$I = -[x]_{-1}^{0} + [x]_{0}^{2}$
$I = -(0 - (-1)) + (2 - 0)$
$I = -(1) + 2$
$I = 1$.

(D) The integral $\int_{0}^{11} [x] dx$ can be split into intervals of unit length where $[x]$ is constant:
$\int_{0}^{11} [x] dx = \sum_{k=0}^{10} \int_{k}^{k+1} [x] dx$
Since for $x \in [k, k+1)$,$[x] = k$,we have:
$\int_{k}^{k+1} [x] dx = \int_{k}^{k+1} k dx = k(k+1 - k) = k$
Therefore,the sum becomes:
$\int_{0}^{11} [x] dx = \sum_{k=0}^{10} k = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10$
Using the sum formula for the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:
$\sum_{k=1}^{10} k = \frac{10 \times 11}{2} = 55$
Thus,the value of the integral is $55$.

(B) To evaluate $\int_{0}^{4}|x-1| dx$,we first identify the point where the expression inside the modulus changes sign,which is $x = 1$.
We split the integral at $x = 1$:
$\int_{0}^{4}|x-1| dx = \int_{0}^{1}-(x-1) dx + \int_{1}^{4}(x-1) dx$
Evaluating the first part:
$\int_{0}^{1}(-x+1) dx = [-\frac{x^2}{2} + x]_{0}^{1} = (-\frac{1}{2} + 1) - (0) = \frac{1}{2}$
Evaluating the second part:
$\int_{1}^{4}(x-1) dx = [\frac{x^2}{2} - x]_{1}^{4} = (\frac{16}{2} - 4) - (\frac{1}{2} - 1) = (8 - 4) - (-\frac{1}{2}) = 4 + \frac{1}{2} = \frac{9}{2}$
Adding both parts:
$\frac{1}{2} + \frac{9}{2} = \frac{10}{2} = 5$

(C) Consider the given integral $I = \int_{0}^{2} [x^{2}] \, dx$.
Since the greatest integer function $[x^2]$ changes its value at $x^2 = 1, 2, 3$,we split the interval $[0, 2]$ based on $x = 1, \sqrt{2}, \sqrt{3}$.
For $x \in [0, 1)$,$x^2 \in [0, 1)$,so $[x^2] = 0$.
For $x \in [1, \sqrt{2})$,$x^2 \in [1, 2)$,so $[x^2] = 1$.
For $x \in [\sqrt{2}, \sqrt{3})$,$x^2 \in [2, 3)$,so $[x^2] = 2$.
For $x \in [\sqrt{3}, 2]$,$x^2 \in [3, 4]$,so $[x^2] = 3$.
Thus,$I = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{2} 3 \, dx$.
$I = 0 + [x]_{1}^{\sqrt{2}} + [2x]_{\sqrt{2}}^{\sqrt{3}} + [3x]_{\sqrt{3}}^{2}$.
$I = (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$.
$I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$.
$I = 5 - \sqrt{2} - \sqrt{3}$.

(B) Given,$f(x) = \int_{-1}^{x} |t| dt$.
Since $x \geq 0$,we can split the integral at $t = 0$:
$f(x) = \int_{-1}^{0} |t| dt + \int_{0}^{x} |t| dt$.
For $t \in [-1, 0]$,$|t| = -t$,and for $t \in [0, x]$,$|t| = t$.
So,$f(x) = \int_{-1}^{0} (-t) dt + \int_{0}^{x} t dt$.
Evaluating the integrals:
$f(x) = -\left[ \frac{t^{2}}{2} \right]_{-1}^{0} + \left[ \frac{t^{2}}{2} \right]_{0}^{x}$.
$f(x) = -\left( 0 - \frac{(-1)^{2}}{2} \right) + \left( \frac{x^{2}}{2} - 0 \right)$.
$f(x) = -\left( -\frac{1}{2} \right) + \frac{x^{2}}{2} = \frac{1}{2} + \frac{x^{2}}{2} = \frac{1}{2}(1 + x^{2})$.

(A) Given integral is $I_{n} = \int_{0}^{\pi / 4} \tan ^{n} x d x$.
We can write $\tan^{n} x$ as $\tan^{n-2} x \cdot \tan^{2} x$.
Using the identity $\tan^{2} x = \sec^{2} x - 1$,we have:
$I_{n} = \int_{0}^{\pi / 4} \tan^{n-2} x (\sec^{2} x - 1) d x$.
$I_{n} = \int_{0}^{\pi / 4} \tan^{n-2} x \sec^{2} x d x - \int_{0}^{\pi / 4} \tan^{n-2} x d x$.
Let $t = \tan x$,then $dt = \sec^{2} x dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$.
$I_{n} = \int_{0}^{1} t^{n-2} dt - I_{n-2}$.
$I_{n} = \left[ \frac{t^{n-1}}{n-1} \right]_{0}^{1} - I_{n-2}$.
$I_{n} = \frac{1}{n-1} - I_{n-2}$.
Therefore,$I_{n} + I_{n-2} = \frac{1}{n-1}$.
For $n=10$,we get $I_{10} + I_{8} = \frac{1}{10-1} = \frac{1}{9}$.

(B) We are given the definite integral $\int_{0.2}^{3.5} [x] \, dx$,where $[x]$ denotes the greatest integer function.
We split the interval $[0.2, 3.5]$ based on the points where the greatest integer function changes its value:
$\int_{0.2}^{3.5} [x] \, dx = \int_{0.2}^{1} [x] \, dx + \int_{1}^{2} [x] \, dx + \int_{2}^{3} [x] \, dx + \int_{3}^{3.5} [x] \, dx$
Since $[x] = 0$ for $x \in [0.2, 1)$,$[x] = 1$ for $x \in [1, 2)$,$[x] = 2$ for $x \in [2, 3)$,and $[x] = 3$ for $x \in [3, 3.5]$,we have:
$= \int_{0.2}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{3.5} 3 \, dx$
$= 0 + [x]_{1}^{2} + 2[x]_{2}^{3} + 3[x]_{3}^{3.5}$
$= 0 + (2 - 1) + 2(3 - 2) + 3(3.5 - 3)$
$= 0 + 1 + 2(1) + 3(0.5)$
$= 1 + 2 + 1.5 = 4.5$

(A) Given equation: $5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}$ ...$(i)$
Replace $x$ with $\frac{1}{x}$ in equation $(i)$:
$5 f\left(\frac{1}{x}\right)+3 f(x)=2-x$ ...(ii)
To eliminate $f(x)$,multiply equation (ii) by $5$ and equation $(i)$ by $3$:
$25 f\left(\frac{1}{x}\right)+15 f(x)=10-5 x$ ...(iii)
$15 f(x)+9 f\left(\frac{1}{x}\right)=6-\frac{3}{x}$ ...(iv)
Subtract equation (iv) from equation (iii):
$(25-9) f\left(\frac{1}{x}\right) = (10-5 x) - (6-\frac{3}{x})$
$16 f\left(\frac{1}{x}\right) = 4-5 x+\frac{3}{x}$
$f\left(\frac{1}{x}\right) = \frac{1}{16} \left(4-5 x+\frac{3}{x}\right)$
Now,calculate the integral:
$\int_1^2 f\left(\frac{1}{x}\right) d x = \int_1^2 \frac{1}{16} \left(4-5 x+\frac{3}{x}\right) d x$
$= \frac{1}{16} \left[4 x-\frac{5 x^2}{2}+3 \ln |x|\right]_1^2$
$= \frac{1}{16} \left[ \left(4(2)-\frac{5(2)^2}{2}+3 \ln 2\right) - \left(4(1)-\frac{5(1)^2}{2}+3 \ln 1\right) \right]$
$= \frac{1}{16} \left[ (8-10+3 \ln 2) - (4-2.5+0) \right]$
$= \frac{1}{16} \left[ (-2+3 \ln 2) - 1.5 \right]$
$= \frac{1}{16} \left[ 3 \ln 2 - 3.5 \right] = \frac{1}{16} \left[ 3 \ln 2 - \frac{7}{2} \right]$
$= \frac{6 \ln 2-7}{32}$

(C) We are given the integral $I = \int_{1}^{2} \frac{x^{3} - 1}{x^{2}} dx$.
First,simplify the integrand:
$\frac{x^{3} - 1}{x^{2}} = \frac{x^{3}}{x^{2}} - \frac{1}{x^{2}} = x - x^{-2}$.
Now,integrate term by term:
$I = \int_{1}^{2} (x - x^{-2}) dx = \left[ \frac{x^{2}}{2} - \frac{x^{-1}}{-1} \right]_{1}^{2} = \left[ \frac{x^{2}}{2} + \frac{1}{x} \right]_{1}^{2}$.
Substitute the limits:
$I = \left( \frac{2^{2}}{2} + \frac{1}{2} \right) - \left( \frac{1^{2}}{2} + \frac{1}{1} \right) = \left( 2 + \frac{1}{2} \right) - \left( \frac{1}{2} + 1 \right) = 2 + \frac{1}{2} - \frac{1}{2} - 1 = 1$.

(A) If a function $f$ is integrable on $[0, a]$,then any function derived from it by a linear transformation of the variable,such as $f(a-x)$,is also integrable on $[0, a]$.
This is a standard property of Riemann integrable functions.
Specifically,if $f$ is integrable on $[0, a]$,then the composition $f(a-x)$ is integrable on $[0, a]$ because the transformation $u = a-x$ is a continuous and monotonic mapping.
Therefore,$f(a-x)$ is integrable on $[0, a]$.
Thus,option $A$ is correct.

(C) Expand the integrand: $(\sqrt{k}-\sqrt{t})^2 = k + t - 2\sqrt{k}\sqrt{t}$.
Now,integrate term by term with respect to $t$ from $0$ to $k$:
$\int_0^k (k + t - 2\sqrt{k}t^{1/2}) \, dt = \left[ kt + \frac{t^2}{2} - 2\sqrt{k} \cdot \frac{t^{3/2}}{3/2} \right]_0^k$
$= \left[ kt + \frac{t^2}{2} - \frac{4}{3}\sqrt{k}t^{3/2} \right]_0^k$
Substitute the limits:
$= (k(k) + \frac{k^2}{2} - \frac{4}{3}\sqrt{k}(k^{3/2})) - (0)$
$= k^2 + \frac{k^2}{2} - \frac{4}{3}k^2$
$= \frac{3k^2}{2} - \frac{4k^2}{3} = \frac{9k^2 - 8k^2}{6} = \frac{k^2}{6}$.
Thus,the correct option is $C$.

(C) We need to evaluate the integral $I = \int_0^{\pi / 4} \tan^2(x) \, dx$.
Using the trigonometric identity $\tan^2(x) = \sec^2(x) - 1$,we can rewrite the integral as:
$I = \int_0^{\pi / 4} (\sec^2(x) - 1) \, dx$.
Now,integrate each term separately:
$I = [\tan(x) - x]_0^{\pi / 4}$.
Applying the limits:
$I = (\tan(\frac{\pi}{4}) - \frac{\pi}{4}) - (\tan(0) - 0)$.
Since $\tan(\frac{\pi}{4}) = 1$ and $\tan(0) = 0$,we get:
$I = (1 - \frac{\pi}{4}) - (0 - 0) = 1 - \frac{\pi}{4}$.
Thus,the correct option is $C$.

(C) If $f$ and $g$ are integrable on the closed interval $[a, b]$,then by the linearity property of the definite integral,their sum $f+g$ is also integrable on the same interval $[a, b]$.
Specifically,for any $x_1, x_2 \in [a, b]$,the integrals $\int_{x_1}^{x_2} f(x) dx$ and $\int_{x_1}^{x_2} g(x) dx$ exist.
Therefore,the integral of the sum is the sum of the integrals: $\int_{x_1}^{x_2} (f+g)(x) dx = \int_{x_1}^{x_2} f(x) dx + \int_{x_1}^{x_2} g(x) dx$.
Since this holds for all $x_1, x_2 \in [a, b]$,$f+g$ is integrable on $[a, b]$.
Thus,option $C$ is correct.

(B) Given the integral: $\int_{0}^{a} \frac{dx}{4 + x^2} = \frac{\pi}{8}$
We know that $\int \frac{dx}{k^2 + x^2} = \frac{1}{k} \tan^{-1} \left( \frac{x}{k} \right) + C$.
Here,$k^2 = 4$,so $k = 2$.
Applying the limits from $0$ to $a$:
$\left[ \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) \right]_{0}^{a} = \frac{\pi}{8}$
$\frac{1}{2} \tan^{-1} \left( \frac{a}{2} \right) - \frac{1}{2} \tan^{-1} (0) = \frac{\pi}{8}$
Since $\tan^{-1}(0) = 0$,we have:
$\frac{1}{2} \tan^{-1} \left( \frac{a}{2} \right) = \frac{\pi}{8}$
$\tan^{-1} \left( \frac{a}{2} \right) = \frac{\pi}{4}$
Taking the tangent of both sides:
$\frac{a}{2} = \tan \left( \frac{\pi}{4} \right)$
$\frac{a}{2} = 1$
$a = 2$

(B) Given the integral $\int_1^n [x] dx = 120$.
We can split the integral into intervals of unit length:
$\int_1^2 1 dx + \int_2^3 2 dx + \int_3^4 3 dx + \dots + \int_{n-1}^n (n-1) dx = 120$.
This simplifies to the sum of the first $(n-1)$ natural numbers:
$1 + 2 + 3 + \dots + (n-1) = 120$.
Using the formula for the sum of the first $k$ natural numbers,$\frac{k(k+1)}{2}$,where $k = n-1$:
$\frac{(n-1)n}{2} = 120$.
$(n-1)n = 240$.
$n^2 - n - 240 = 0$.
$(n - 16)(n + 15) = 0$.
Since $n$ must be positive,$n = 16$.

(A) Let $I = \int_{\pi / 4}^{\pi / 3} \frac{\cos x-\sin x}{\sin 2 x} d x$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 1 - (\sin x - \cos x)^2$.
Let $u = \sin x - \cos x$. Then $du = (\cos x + \sin x) dx$. This does not simplify directly.
Alternatively,split the integral: $\int \frac{\cos x}{\sin 2x} dx - \int \frac{\sin x}{\sin 2x} dx = \int \frac{\cos x}{2 \sin x \cos x} dx - \int \frac{\sin x}{2 \sin x \cos x} dx = \frac{1}{2} \int \csc x dx - \frac{1}{2} \int \sec x dx$.
$I = \frac{1}{2} [\log |\tan(x/2)| - \log |\sec x + \tan x|]_{\pi/4}^{\pi/3} = \frac{1}{2} [\log |\frac{\tan(x/2)}{\sec x + \tan x}|]_{\pi/4}^{\pi/3}$.
At $x = \pi/3$: $\tan(\pi/6) = 1/\sqrt{3}$,$\sec(\pi/3) = 2$,$\tan(\pi/3) = \sqrt{3}$. Value $= \frac{1/\sqrt{3}}{2+\sqrt{3}} = \frac{1}{\sqrt{3}(2+\sqrt{3})} = \frac{2-\sqrt{3}}{\sqrt{3}}$.
At $x = \pi/4$: $\tan(\pi/8) = \sqrt{2}-1$,$\sec(\pi/4) = \sqrt{2}$,$\tan(\pi/4) = 1$. Value $= \frac{\sqrt{2}-1}{\sqrt{2}+1} = (\sqrt{2}-1)^2 = 3-2\sqrt{2}$.
Thus,$I = \frac{1}{2} \log \left( \frac{(2-\sqrt{3})/\sqrt{3}}{3-2\sqrt{2}} \right) = \frac{1}{2} \log \left( \frac{(2-\sqrt{3})(3+2\sqrt{2})}{\sqrt{3}} \right)$.
This matches option $A$.

(D) We need to evaluate the integral $I = \int_{\pi / 6}^{\pi / 3} \cos^{-4} x \, dx = \int_{\pi / 6}^{\pi / 3} \sec^4 x \, dx$.
Using the identity $\sec^2 x = 1 + \tan^2 x$,we can write $\sec^4 x = \sec^2 x \cdot \sec^2 x = (1 + \tan^2 x) \sec^2 x$.
So,$I = \int_{\pi / 6}^{\pi / 3} (1 + \tan^2 x) \sec^2 x \, dx$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
When $x = \pi / 6$,$u = \tan(\pi / 6) = 1 / \sqrt{3}$.
When $x = \pi / 3$,$u = \tan(\pi / 3) = \sqrt{3}$.
Substituting these into the integral:
$I = \int_{1 / \sqrt{3}}^{\sqrt{3}} (1 + u^2) \, du = [u + \frac{u^3}{3}]_{1 / \sqrt{3}}^{\sqrt{3}}$.
Evaluating at the limits:
$I = (\sqrt{3} + \frac{(\sqrt{3})^3}{3}) - (\frac{1}{\sqrt{3}} + \frac{(1 / \sqrt{3})^3}{3}) = (\sqrt{3} + \sqrt{3}) - (\frac{1}{\sqrt{3}} + \frac{1}{9 \sqrt{3}}) = 2 \sqrt{3} - (\frac{9 + 1}{9 \sqrt{3}}) = 2 \sqrt{3} - \frac{10}{9 \sqrt{3}}$.
To combine,$I = \frac{2 \sqrt{3} \cdot 9 \sqrt{3} - 10}{9 \sqrt{3}} = \frac{2 \cdot 9 \cdot 3 - 10}{9 \sqrt{3}} = \frac{54 - 10}{9 \sqrt{3}} = \frac{44}{9 \sqrt{3}}$.
Thus,the correct option is $D$.

(A) To evaluate the integral $I = \int_0^1 x \operatorname{Sin}^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \operatorname{Sin}^{-1} x$ and $dv = x \, dx$.
Then $du = \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = \frac{x^2}{2}$.
Applying the formula:
$I = \left[ \frac{x^2}{2} \operatorname{Sin}^{-1} x \right]_0^1 - \int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} \, dx$.
Evaluating the first part: $\left[ \frac{1^2}{2} \operatorname{Sin}^{-1}(1) - 0 \right] = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$.
For the second part,let $J = \int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} \, dx$. Substitute $x = \sin \theta$,$dx = \cos \theta \, d\theta$.
$J = \frac{1}{2} \int_0^{\pi/2} \frac{\sin^2 \theta \cos \theta}{\cos \theta} \, d\theta = \frac{1}{2} \int_0^{\pi/2} \sin^2 \theta \, d\theta$.
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:
$J = \frac{1}{4} \int_0^{\pi/2} (1 - \cos 2\theta) \, d\theta = \frac{1}{4} \left[ \theta - \frac{\sin 2\theta}{2} \right]_0^{\pi/2} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}$.
Thus,$I = \frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8}$.

(B) Let $I = \int \frac{x^2}{(x \sin x + \cos x)^2} dx$.
We can rewrite the integrand as $I = \int (x \sec x) \left( \frac{x \cos x}{(x \sin x + \cos x)^2} \right) dx$.
Using integration by parts,let $u = x \sec x$ and $dv = \frac{x \cos x}{(x \sin x + \cos x)^2} dx$.
Then $du = (\sec x + x \sec x \tan x) dx$ and $v = \frac{-1}{x \sin x + \cos x}$.
$I = (x \sec x) \left( \frac{-1}{x \sin x + \cos x} \right) - \int (\sec x + x \sec x \tan x) \left( \frac{-1}{x \sin x + \cos x} \right) dx$.
$I = \frac{-x \sec x}{x \sin x + \cos x} + \int \frac{\sec x (1 + x \tan x)}{x \sin x + \cos x} dx$.
Since $x \sin x + \cos x = \cos x (x \tan x + 1)$,the integral becomes $\int \frac{\sec x \cdot \cos x (x \tan x + 1)}{\cos x (x \tan x + 1)} dx = \int \sec^2 x dx = \tan x$.
Thus,$I = \frac{-x \sec x}{x \sin x + \cos x} + \tan x = \frac{-x + \tan x (x \sin x + \cos x)}{\cos x (x \sin x + \cos x)} = \frac{\sin x - x \cos x}{x \sin x + \cos x}$.
Evaluating the definite integral: $\left[ \frac{\sin x - x \cos x}{x \sin x + \cos x} \right]_0^{\pi / 4} = \frac{\frac{1}{\sqrt{2}} - \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}}{\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} - 0 = \frac{1 - \pi/4}{1 + \pi/4} = \frac{4 - \pi}{4 + \pi}$.

(D) Let $I = \int_0^1 (1+x) \log (1+x) \, dx$.
Using integration by parts,let $u = \log(1+x)$ and $dv = (1+x) \, dx$.
Then $du = \frac{1}{1+x} \, dx$ and $v = \frac{(1+x)^2}{2}$.
$I = \left[ \frac{(1+x)^2}{2} \log(1+x) \right]_0^1 - \int_0^1 \frac{(1+x)^2}{2} \cdot \frac{1}{1+x} \, dx$
$I = \left[ \frac{4}{2} \log 2 - 0 \right] - \frac{1}{2} \int_0^1 (1+x) \, dx$
$I = 2 \log 2 - \frac{1}{2} \left[ x + \frac{x^2}{2} \right]_0^1$
$I = 2 \log 2 - \frac{1}{2} \left( 1 + \frac{1}{2} \right) = 2 \log 2 - \frac{1}{2} \left( \frac{3}{2} \right)$
$I = 2 \log 2 - \frac{3}{4} = \frac{-3}{4} + 2 \log 2$.

(C) Let $I = \int_{0}^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx$.
Using the identities $1 + \cos x = 2 \cos^{2} \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have:
$I = \int_{0}^{\pi/2} \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} dx$
$I = \int_{0}^{\pi/2} \left( \frac{x}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2} \right) dx$
Using integration by parts for the first term $\int \frac{x}{2} \sec^{2} \frac{x}{2} dx$:
Let $u = \frac{x}{2}$,$dv = \sec^{2} \frac{x}{2} dx$. Then $du = \frac{1}{2} dx$ and $v = 2 \tan \frac{x}{2}$.
$\int \frac{x}{2} \sec^{2} \frac{x}{2} dx = \frac{x}{2} (2 \tan \frac{x}{2}) - \int (2 \tan \frac{x}{2}) \cdot \frac{1}{2} dx = x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx$.
Substituting this back into the integral $I$:
$I = [x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx] + \int \tan \frac{x}{2} dx$
$I = [x \tan \frac{x}{2}]_{0}^{\pi/2}$
$I = \frac{\pi}{2} \tan \frac{\pi}{4} - 0 \cdot \tan 0 = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2}$.

(B) We need to evaluate $I = \int_{-1}^{3/2} |x \sin \pi x| \, dx$.
Since $x \sin \pi x \ge 0$ for $x \in [-1, 0]$ and $x \sin \pi x \ge 0$ for $x \in [0, 1]$,but $x \sin \pi x \le 0$ for $x \in [1, 3/2]$,we split the integral:
$I = \int_{-1}^{1} x \sin \pi x \, dx - \int_{1}^{3/2} x \sin \pi x \, dx$.
Using integration by parts $\int u \, dv = uv - \int v \, du$,let $u = x$ and $dv = \sin \pi x \, dx$,so $du = dx$ and $v = -\frac{\cos \pi x}{\pi}$.
$\int x \sin \pi x \, dx = -\frac{x \cos \pi x}{\pi} + \frac{1}{\pi} \int \cos \pi x \, dx = -\frac{x \cos \pi x}{\pi} + \frac{\sin \pi x}{\pi^2}$.
Evaluating the first part: $\int_{-1}^{1} x \sin \pi x \, dx = [-\frac{x \cos \pi x}{\pi} + \frac{\sin \pi x}{\pi^2}]_{-1}^{1} = (-\frac{1 \cdot (-1)}{\pi} + 0) - (-\frac{-1 \cdot (-1)}{\pi} + 0) = \frac{1}{\pi} - \frac{1}{\pi} = 0$ is incorrect; let's re-evaluate:
$[-\frac{1 \cdot (-1)}{\pi} + 0] - [-\frac{-1 \cdot (-1)}{\pi} + 0] = \frac{1}{\pi} - (-\frac{1}{\pi}) = \frac{2}{\pi}$.
Evaluating the second part: $\int_{1}^{3/2} x \sin \pi x \, dx = [-\frac{x \cos \pi x}{\pi} + \frac{\sin \pi x}{\pi^2}]_{1}^{3/2} = (0 + \frac{\sin(3\pi/2)}{\pi^2}) - (-\frac{1 \cdot (-1)}{\pi} + 0) = -\frac{1}{\pi^2} - \frac{1}{\pi}$.
Thus,$I = \frac{2}{\pi} - (-\frac{1}{\pi^2} - \frac{1}{\pi}) = \frac{2}{\pi} + \frac{1}{\pi} + \frac{1}{\pi^2} = \frac{3}{\pi} + \frac{1}{\pi^2}$.

(C) Given,$I_n = \int_0^{\pi/4} \tan^n \theta \, d\theta$.
We need to find $I_{n-1} + I_{n+1}$.
$I_{n-1} + I_{n+1} = \int_0^{\pi/4} \tan^{n-1} \theta \, d\theta + \int_0^{\pi/4} \tan^{n+1} \theta \, d\theta$.
Factor out $\tan^{n-1} \theta$:
$I_{n-1} + I_{n+1} = \int_0^{\pi/4} \tan^{n-1} \theta (1 + \tan^2 \theta) \, d\theta$.
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have:
$I_{n-1} + I_{n+1} = \int_0^{\pi/4} \tan^{n-1} \theta \sec^2 \theta \, d\theta$.
Let $u = \tan \theta$,then $du = \sec^2 \theta \, d\theta$.
When $\theta = 0$,$u = 0$. When $\theta = \pi/4$,$u = 1$.
$I_{n-1} + I_{n+1} = \int_0^1 u^{n-1} \, du = \left[ \frac{u^n}{n} \right]_0^1 = \frac{1}{n} - 0 = \frac{1}{n}$.

(A) Let $I = \int_0^2 \frac{2x-2}{2x-x^2} dx$.
Note that the integrand $f(x) = \frac{2x-2}{2x-x^2} = \frac{-(2-2x)}{x(2-x)} = \frac{2(x-1)}{x(2-x)}$.
The integral is improper because the denominator $2x-x^2 = x(2-x)$ becomes $0$ at $x=0$ and $x=2$.
Let us evaluate the indefinite integral: $\int \frac{2x-2}{2x-x^2} dx = \int \frac{-(2-2x)}{2x-x^2} dx$.
Let $u = 2x-x^2$,then $du = (2-2x) dx$,so $-(2-2x) dx = -du$.
Thus,$\int \frac{-du}{u} = -\ln|u| + C = -\ln|2x-x^2| + C$.
Evaluating the definite integral as a limit: $\lim_{\epsilon \to 0^+} \int_{\epsilon}^{2-\epsilon} \frac{2x-2}{2x-x^2} dx = \lim_{\epsilon \to 0^+} [-\ln|2x-x^2|]_{\epsilon}^{2-\epsilon}$.
$= \lim_{\epsilon \to 0^+} [-\ln|2(2-\epsilon)-(2-\epsilon)^2| - (-\ln|2\epsilon-\epsilon^2|)]$.
$= \lim_{\epsilon \to 0^+} [-\ln|4-2\epsilon-(4-4\epsilon+\epsilon^2)| + \ln|2\epsilon-\epsilon^2|]$.
$= \lim_{\epsilon \to 0^+} [-\ln|2\epsilon-\epsilon^2| + \ln|2\epsilon-\epsilon^2|] = 0$.
However,since the integral diverges at both endpoints,the Cauchy Principal Value is $0$.

(D) Let $I = \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} \, dx$.
Substitute $x = 4 \cos^2 \theta - 1 \sin^2 \theta$ is not ideal. Let $x+1 = 5 \cos^2 \theta$,then $dx = -10 \cos \theta \sin \theta \, d\theta$.
When $x = -1$,$5 \cos^2 \theta = 0 \implies \theta = \frac{\pi}{2}$.
When $x = 4$,$5 \cos^2 \theta = 5 \implies \theta = 0$.
Then $4-x = 4 - (5 \cos^2 \theta - 1) = 5 - 5 \cos^2 \theta = 5 \sin^2 \theta$.
$I = \int_{\pi/2}^0 \sqrt{\frac{5 \sin^2 \theta}{5 \cos^2 \theta}} (-10 \cos \theta \sin \theta) \, d\theta = \int_0^{\pi/2} \tan \theta (10 \cos \theta \sin \theta) \, d\theta$.
$I = 10 \int_0^{\pi/2} \sin^2 \theta \, d\theta = 10 \int_0^{\pi/2} \frac{1 - \cos 2\theta}{2} \, d\theta$.
$I = 5 [\theta - \frac{\sin 2\theta}{2}]_0^{\pi/2} = 5 [\frac{\pi}{2} - 0] = \frac{5\pi}{2}$.