Fundamental definite integration Questions in English

(D) Let $I = \int_0^{\pi / 4} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} dx$.
Divide numerator and denominator by $\cos^2 x$:
$I = \int_0^{\pi / 4} \frac{1}{1 + 4 \tan^2 x} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx = (1 + u^2) dx$,so $dx = \frac{du}{1 + u^2}$.
When $x = 0, u = 0$. When $x = \pi / 4, u = 1$.
$I = \int_0^1 \frac{1}{(1 + 4u^2)(1 + u^2)} du$.
Using partial fractions: $\frac{1}{(1 + 4u^2)(1 + u^2)} = \frac{A}{1 + 4u^2} + \frac{B}{1 + u^2}$.
$1 = A(1 + u^2) + B(1 + 4u^2)$.
For $u^2 = -1, 1 = B(1 - 4) \implies B = -1/3$.
For $u^2 = -1/4, 1 = A(1 - 1/4) \implies A = 4/3$.
$I = \int_0^1 (\frac{4/3}{1 + 4u^2} - \frac{1/3}{1 + u^2}) du$.
$I = \frac{4}{3} \int_0^1 \frac{1}{1 + (2u)^2} du - \frac{1}{3} \int_0^1 \frac{1}{1 + u^2} du$.
$I = \frac{4}{3} [\frac{1}{2} \tan^{-1}(2u)]_0^1 - \frac{1}{3} [\tan^{-1} u]_0^1$.
$I = \frac{2}{3} \tan^{-1} 2 - \frac{1}{3} (\frac{\pi}{4}) = \frac{2}{3} \tan^{-1} 2 - \frac{\pi}{12}$.

(D) To evaluate $I = \int_0^{x} \frac{t^2}{\sqrt{a^2+t^2}} dt$,we rewrite the numerator as $t^2 = (t^2+a^2) - a^2$.
Then,$I = \int_0^{x} \frac{t^2+a^2}{\sqrt{a^2+t^2}} dt - \int_0^{x} \frac{a^2}{\sqrt{a^2+t^2}} dt$.
$I = \int_0^{x} \sqrt{a^2+t^2} dt - a^2 \int_0^{x} \frac{1}{\sqrt{a^2+t^2}} dt$.
Using the standard integrals $\int \sqrt{a^2+t^2} dt = \frac{t}{2}\sqrt{a^2+t^2} + \frac{a^2}{2} \log|t+\sqrt{a^2+t^2}|$ and $\int \frac{1}{\sqrt{a^2+t^2}} dt = \log|t+\sqrt{a^2+t^2}|$:
$I = \left[ \frac{t}{2}\sqrt{a^2+t^2} + \frac{a^2}{2} \log|t+\sqrt{a^2+t^2}| - a^2 \log|t+\sqrt{a^2+t^2}| \right]_0^x$.
$I = \left[ \frac{t}{2}\sqrt{a^2+t^2} - \frac{a^2}{2} \log|t+\sqrt{a^2+t^2}| \right]_0^x$.
Evaluating at limits: $I = \left( \frac{x}{2}\sqrt{a^2+x^2} - \frac{a^2}{2} \log|x+\sqrt{a^2+x^2}| \right) - \left( 0 - \frac{a^2}{2} \log|a| \right)$.
$I = \frac{x}{2}\sqrt{a^2+x^2} - \frac{a^2}{2} \log \left| \frac{x+\sqrt{a^2+x^2}}{a} \right|$.

(A) To evaluate $\int_0^2 f(x) dx$,we split the integral at $x = 1$:
$\int_0^2 f(x) dx = \int_0^1 \frac{6 x^2 + 1}{4 x^3 + 2 x + 3} dx + \int_1^2 (x^2 + 1) dx$
For the first integral,let $u = 4 x^3 + 2 x + 3$,then $du = (12 x^2 + 2) dx = 2(6 x^2 + 1) dx$.
Thus,$\int_0^1 \frac{6 x^2 + 1}{4 x^3 + 2 x + 3} dx = \frac{1}{2} \int_{u(0)}^{u(1)} \frac{1}{u} du = \frac{1}{2} [\log |u|]_3^9 = \frac{1}{2} (\log 9 - \log 3) = \frac{1}{2} \log 3$.
For the second integral,$\int_1^2 (x^2 + 1) dx = [\frac{x^3}{3} + x]_1^2 = (\frac{8}{3} + 2) - (\frac{1}{3} + 1) = \frac{14}{3} - \frac{4}{3} = \frac{10}{3}$.
Adding both parts,$\int_0^2 f(x) dx = \frac{1}{2} \log 3 + \frac{10}{3}$.

(D) Let $I = \int_0^1 \frac{x}{(1-x)^{3/4}} dx$.
Substitute $t = 1 - x$,then $dt = -dx$. When $x = 0, t = 1$ and when $x = 1, t = 0$.
$I = \int_1^0 \frac{1-t}{t^{3/4}} (-dt) = \int_0^1 \frac{1-t}{t^{3/4}} dt$.
$I = \int_0^1 (t^{-3/4} - t^{1/4}) dt$.
$I = \left[ \frac{t^{1/4}}{1/4} - \frac{t^{5/4}}{5/4} \right]_0^1$.
$I = \left[ 4t^{1/4} - \frac{4}{5}t^{5/4} \right]_0^1 = 4 - \frac{4}{5} = \frac{20-4}{5} = \frac{16}{5}$.

(D) Let $I = \int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx$.
Rationalizing the integrand: $I = \int_0^1 \frac{2+x}{\sqrt{4-x^2}} \, dx = \int_0^1 \frac{2}{\sqrt{4-x^2}} \, dx + \int_0^1 \frac{x}{\sqrt{4-x^2}} \, dx$.
For the first part,$\int_0^1 \frac{2}{\sqrt{2^2-x^2}} \, dx = 2 \left[ \sin^{-1} \left( \frac{x}{2} \right) \right]_0^1 = 2 \left( \sin^{-1} \frac{1}{2} - \sin^{-1} 0 \right) = 2 \left( \frac{\pi}{6} - 0 \right) = \frac{\pi}{3}$.
For the second part,let $t = 4-x^2$,then $dt = -2x \, dx$,so $x \, dx = -\frac{1}{2} \, dt$.
$\int_0^1 \frac{x}{\sqrt{4-x^2}} \, dx = -\frac{1}{2} \int_4^3 \frac{1}{\sqrt{t}} \, dt = \frac{1}{2} \int_3^4 t^{-1/2} \, dt = \frac{1}{2} [2\sqrt{t}]_3^4 = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3}$.
Adding both parts: $I = \frac{\pi}{3} + 2 - \sqrt{3}$.

(B) Let $I = \int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1-x^2}} dx$.
Substitute $\sin^{-1} x = t$,then $x = \sin t$ and $dx = \cos t dt$.
When $x = 0$,$t = 0$. When $x = 1/2$,$t = \pi/6$.
Substituting these into the integral:
$I = \int_0^{\pi/6} \frac{\sin t \cdot t}{\sqrt{1-\sin^2 t}} \cdot \cos t dt = \int_0^{\pi/6} t \sin t dt$.
Using integration by parts: $\int u dv = uv - \int v du$.
Let $u = t$ and $dv = \sin t dt$,then $du = dt$ and $v = -\cos t$.
$I = [t(-\cos t)]_0^{\pi/6} - \int_0^{\pi/6} (-\cos t) dt$
$I = [-t \cos t + \sin t]_0^{\pi/6}$
$I = [-\frac{\pi}{6} \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6})] - [0 + \sin(0)]$
$I = [-\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}] = \frac{1}{2} - \frac{\sqrt{3}}{12} \pi$.

(C) Given $u(n) = \int_0^{\pi/2} (1 + \sin t)^n \sin 2t \, dt$.
Using the identity $\sin 2t = 2 \sin t \cos t$,we have:
$u(n) = 2 \int_0^{\pi/2} (1 + \sin t)^n \sin t \cos t \, dt$.
Let $x = 1 + \sin t$,then $dx = \cos t \, dt$.
When $t = 0$,$x = 1$. When $t = \pi/2$,$x = 2$.
Substituting these into the integral:
$u(n) = 2 \int_1^2 x^n (x - 1) \, dx = 2 \int_1^2 (x^{n+1} - x^n) \, dx$.
Integrating,we get:
$u(n) = 2 \left[ \frac{x^{n+2}}{n+2} - \frac{x^{n+1}}{n+1} \right]_1^2$.
For $n = 4$:
$u(4) = 2 \left[ \frac{x^6}{6} - \frac{x^5}{5} \right]_1^2 = 2 \left( (\frac{2^6}{6} - \frac{2^5}{5}) - (\frac{1}{6} - \frac{1}{5}) \right)$.
$u(4) = 2 \left( (\frac{64}{6} - \frac{32}{5}) - (\frac{5 - 6}{30}) \right) = 2 \left( \frac{320 - 192}{30} + \frac{1}{30} \right)$.
$u(4) = 2 \left( \frac{128}{30} + \frac{1}{30} \right) = 2 \left( \frac{129}{30} \right) = \frac{129}{15}$.

(C) Let $I = \int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx$.
Substitute $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$.
When $x = -\pi$,$t = \cos(-\pi) = -1$.
When $x = \frac{\pi}{2}$,$t = \cos(\frac{\pi}{2}) = 0$.
Thus,$I = \int_{-1}^{0} \sin^2(t) (-dt) = \int_{0}^{-1} \sin^2(t) \, dt$.
Using the identity $\sin^2(t) = \frac{1 - \cos(2t)}{2}$,we get:
$I = \int_{0}^{-1} \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \left[ t - \frac{\sin(2t)}{2} \right]_{0}^{-1}$.
$I = \frac{1}{2} \left[ (-1 - \frac{\sin(-2)}{2}) - (0 - 0) \right]$.
Since $\sin(-2) = -\sin(2)$,we have:
$I = \frac{1}{2} \left[ -1 + \frac{\sin(2)}{2} \right] = \frac{\sin(2) - 2}{4}$.

(D) Given the equation $\int_x^1(1-t) dt = \frac{1}{2}$.
Evaluating the integral: $\left[t - \frac{t^2}{2}\right]_x^1 = \frac{1}{2}$.
Substituting the limits: $(1 - \frac{1}{2}) - (x - \frac{x^2}{2}) = \frac{1}{2}$.
$\frac{1}{2} - x + \frac{x^2}{2} = \frac{1}{2}$.
Subtracting $\frac{1}{2}$ from both sides: $\frac{x^2}{2} - x = 0$.
Multiplying by $2$: $x^2 - 2x = 0$.
Factoring: $x(x - 2) = 0$.
Thus,$x = 0$ or $x = 2$.
Since the question asks for the positive value of $x$,we have $x = 2$.

(A) Let $I = \int_{0}^{\frac{\pi}{4}} e^{\tan^2 \theta} \sin^2 \theta \tan \theta d\theta$.
Since $\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta}$,we have $I = \int_{0}^{\frac{\pi}{4}} e^{\tan^2 \theta} \frac{\tan^3 \theta}{1 + \tan^2 \theta} d\theta$.
Let $\tan^2 \theta = u$,then $2 \tan \theta \sec^2 \theta d\theta = du$.
Since $\sec^2 \theta = 1 + \tan^2 \theta = 1 + u$,we have $d\theta = \frac{du}{2 \tan \theta (1 + u)} = \frac{du}{2 \sqrt{u} (1 + u)}$.
Substituting these into the integral:
$I = \int_{0}^{1} e^u \frac{u^{3/2}}{1 + u} \frac{du}{2 \sqrt{u} (1 + u)} = \frac{1}{2} \int_{0}^{1} \frac{u e^u}{(1 + u)^2} du$.
Using integration by parts,let $f(u) = u e^u$ and $g'(u) = (1 + u)^{-2}$,so $f'(u) = (u+1)e^u$ and $g(u) = -(1+u)^{-1}$.
$I = \frac{1}{2} \left[ -\frac{u e^u}{1 + u} \right]_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{(u+1)e^u}{1+u} du$.
$I = \frac{1}{2} \left( -\frac{e}{2} + 0 \right) + \frac{1}{2} \int_{0}^{1} e^u du$.
$I = -\frac{e}{4} + \frac{1}{2} [e^u]_{0}^{1} = -\frac{e}{4} + \frac{1}{2} (e - 1) = \frac{e}{4} - \frac{1}{2} = \frac{1}{2} \left( \frac{e}{2} - 1 \right)$.

(B) Let $I = \int_0^4 \frac{x+2}{\sqrt{4x-x^2}} dx$.
First,rewrite the numerator: $x+2 = \frac{1}{2}(2x-4) + 4$.
So,$I = \int_0^4 \frac{\frac{1}{2}(2x-4) + 4}{\sqrt{4x-x^2}} dx = \frac{1}{2} \int_0^4 \frac{2x-4}{\sqrt{4x-x^2}} dx + 4 \int_0^4 \frac{1}{\sqrt{4-(x-2)^2}} dx$.
For the first integral,let $u = 4x-x^2$,then $du = (4-2x) dx$,so $\int \frac{2x-4}{\sqrt{4x-x^2}} dx = -2\sqrt{4x-x^2}$.
Evaluating from $0$ to $4$: $[-2\sqrt{4x-x^2}]_0^4 = -2(0) - (-2(0)) = 0$.
For the second integral,$\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}(\frac{x}{a})$.
So,$4 \int_0^4 \frac{1}{\sqrt{2^2-(x-2)^2}} dx = 4 [\sin^{-1}(\frac{x-2}{2})]_0^4$.
$= 4 [\sin^{-1}(1) - \sin^{-1}(-1)] = 4 [\frac{\pi}{2} - (-\frac{\pi}{2})] = 4 [\pi] = 4\pi$.

(C) Let $I = \int_2^5 (\sqrt{x+2 \sqrt{x-1}} + \sqrt{x-2 \sqrt{x-1}}) dx$.
We can rewrite the terms inside the square roots as perfect squares:
$x + 2\sqrt{x-1} = (\sqrt{x-1})^2 + 2\sqrt{x-1} + 1 = (\sqrt{x-1} + 1)^2$.
$x - 2\sqrt{x-1} = (\sqrt{x-1})^2 - 2\sqrt{x-1} + 1 = (\sqrt{x-1} - 1)^2$.
Thus,the integral becomes:
$I = \int_2^5 (\sqrt{(\sqrt{x-1} + 1)^2} + \sqrt{(\sqrt{x-1} - 1)^2}) dx$.
Since $x \in [2, 5]$,$\sqrt{x-1} \ge 1$,so $\sqrt{x-1} - 1 \ge 0$.
$I = \int_2^5 (\sqrt{x-1} + 1 + \sqrt{x-1} - 1) dx = \int_2^5 2\sqrt{x-1} dx$.
$I = 2 \int_2^5 (x-1)^{1/2} dx = 2 \left[ \frac{(x-1)^{3/2}}{3/2} \right]_2^5 = 2 \times \frac{2}{3} [(x-1)^{3/2}]_2^5$.
$I = \frac{4}{3} [(5-1)^{3/2} - (2-1)^{3/2}] = \frac{4}{3} [4^{3/2} - 1^{3/2}] = \frac{4}{3} [8 - 1] = \frac{4}{3} \times 7 = \frac{28}{3}$.

(D) Let $I = \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}$.
Expanding the denominator: $(x-a)(b-x) = -x^2 + (a+b)x - ab$.
Completing the square: $-[x^2 - (a+b)x + ab] = -[x^2 - (a+b)x + (\frac{a+b}{2})^2 - (\frac{a+b}{2})^2 + ab] = -[(x - \frac{a+b}{2})^2 - (\frac{b-a}{2})^2] = (\frac{b-a}{2})^2 - (x - \frac{a+b}{2})^2$.
Thus,$I = \int_a^b \frac{dx}{\sqrt{(\frac{b-a}{2})^2 - (x - \frac{a+b}{2})^2}}$.
Using the standard integral $\int \frac{dx}{\sqrt{A^2 - u^2}} = \sin^{-1}(\frac{u}{A}) + C$,we get:
$I = [\sin^{-1}(\frac{x - \frac{a+b}{2}}{\frac{b-a}{2}})]_a^b$.
Evaluating at the limits:
Upper limit $(x=b)$: $\sin^{-1}(\frac{b - \frac{a+b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(\frac{\frac{b-a}{2}}{\frac{b-a}{2}}) = \sin^{-1}(1) = \frac{\pi}{2}$.
Lower limit $(x=a)$: $\sin^{-1}(\frac{a - \frac{a+b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(\frac{\frac{a-b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(-1) = -\frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

(C) Let $I = \int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta d\theta$.
When $x = 0, \theta = 0$ and when $x = 1, \theta = \frac{\pi}{4}$.
Using the identity $\sin^{-1}(\sin 2\theta) = 2\theta$ for $0 \le \theta \le \frac{\pi}{4}$:
$I = \int_0^{\frac{\pi}{4}} 2\theta \sec^2 \theta d\theta = 2 \int_0^{\frac{\pi}{4}} \theta \sec^2 \theta d\theta$.
Using integration by parts: $\int u dv = uv - \int v du$,where $u = \theta$ and $dv = \sec^2 \theta d\theta$.
$I = 2 \left[ \theta \tan \theta - \int \tan \theta d\theta \right]_0^{\frac{\pi}{4}} = 2 [\theta \tan \theta - \log |\sec \theta|]_0^{\frac{\pi}{4}}$.
$I = 2 [(\frac{\pi}{4} \tan \frac{\pi}{4} - \log \sec \frac{\pi}{4}) - (0 - \log \sec 0)]$.
$I = 2 [\frac{\pi}{4}(1) - \log \sqrt{2} - 0] = 2 [\frac{\pi}{4} - \frac{1}{2} \log 2] = \frac{\pi}{2} - \log 2$.

(D) Let $I = \int_0^{\pi/4} (\tan^2 x - \tan^4 x) dx$.
We can factor the integrand as:
$\tan^2 x - \tan^4 x = \tan^2 x (1 - \tan^2 x)$.
Using the identity $\tan^2 x = \sec^2 x - 1$,we have:
$I = \int_0^{\pi/4} (\sec^2 x - 1)(1 - (\sec^2 x - 1)) dx = \int_0^{\pi/4} (\sec^2 x - 1)(2 - \sec^2 x) dx$.
Expanding the product:
$I = \int_0^{\pi/4} (2\sec^2 x - \sec^4 x - 2 + \sec^2 x) dx = \int_0^{\pi/4} (3\sec^2 x - \sec^4 x - 2) dx$.
Using $\sec^4 x = \sec^2 x (1 + \tan^2 x)$:
$I = \int_0^{\pi/4} (3\sec^2 x - \sec^2 x(1 + \tan^2 x) - 2) dx = \int_0^{\pi/4} (2\sec^2 x - \sec^2 x \tan^2 x - 2) dx$.
Integrating term by term:
$I = [2\tan x - \frac{\tan^3 x}{3} - 2x]_0^{\pi/4}$.
Evaluating at the limits:
$I = (2(1) - \frac{1^3}{3} - 2(\frac{\pi}{4})) - (0) = 2 - \frac{1}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{2}$.

(D) $(i)$ Let $I_1 = \int_0^{\pi / 2} \sin ^m x \cos x d x$.
Substitute $\sin x = t$,then $\cos x d x = d t$.
When $x = 0, t = 0$ and when $x = \pi / 2, t = 1$.
$I_1 = \int_0^1 t^m d t = \left[ \frac{t^{m+1}}{m+1} \right]_0^1 = \frac{1}{m+1} (1^{m+1} - 0^{m+1}) = \frac{1}{m+1}$.
Thus,statement $(i)$ is true.
(ii) Let $I_2 = \int_0^{\pi / 2} \sin x \cos ^n x d x$.
Substitute $\cos x = t$,then $-\sin x d x = d t$,or $\sin x d x = -d t$.
When $x = 0, t = 1$ and when $x = \pi / 2, t = 0$.
$I_2 = \int_1^0 t^n (-d t) = \int_0^1 t^n d t = \left[ \frac{t^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1} (1^{n+1} - 0^{n+1}) = \frac{1}{n+1}$.
Thus,statement (ii) is true.
Therefore,both $(i)$ and (ii) are true.

(D) Let $I = \int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx$.
Since $\log x < 0$ for $x \in [e^{-1}, 1)$ and $\log x \ge 0$ for $x \in [1, e^2]$,we split the integral:
$I = \int_{e^{-1}}^{1} \left( -\frac{\log x}{x} \right) dx + \int_{1}^{e^2} \left( \frac{\log x}{x} \right) dx$.
Substitute $t = \log x$,so $dt = \frac{dx}{x}$.
When $x = e^{-1}, t = -1$. When $x = 1, t = 0$. When $x = e^2, t = 2$.
$I = -\int_{-1}^{0} t dt + \int_{0}^{2} t dt$.
$I = -\left[ \frac{t^2}{2} \right]_{-1}^{0} + \left[ \frac{t^2}{2} \right]_{0}^{2}$.
$I = -\left( 0 - \frac{(-1)^2}{2} \right) + \left( \frac{2^2}{2} - 0 \right)$.
$I = -\left( -\frac{1}{2} \right) + \frac{4}{2} = \frac{1}{2} + 2 = \frac{5}{2}$.
Thus,the correct option is $D$.

(A) We need to evaluate the integral $I = \int_{-2}^3 |1-x^2| dx$.
The expression inside the absolute value,$1-x^2$,changes sign at $x = -1$ and $x = 1$.
We split the integral into three intervals: $[-2, -1]$,$[-1, 1]$,and $[1, 3]$.
In $[-2, -1]$,$1-x^2 \le 0$,so $|1-x^2| = x^2-1$.
In $[-1, 1]$,$1-x^2 \ge 0$,so $|1-x^2| = 1-x^2$.
In $[1, 3]$,$1-x^2 \le 0$,so $|1-x^2| = x^2-1$.
Thus,$I = \int_{-2}^{-1} (x^2-1) dx + \int_{-1}^1 (1-x^2) dx + \int_{1}^3 (x^2-1) dx$.
Evaluating each part:
$\int_{-2}^{-1} (x^2-1) dx = [\frac{x^3}{3} - x]_{-2}^{-1} = (-\frac{1}{3} + 1) - (-\frac{8}{3} + 2) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$.
$\int_{-1}^1 (1-x^2) dx = [x - \frac{x^3}{3}]_{-1}^1 = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$.
$\int_{1}^3 (x^2-1) dx = [\frac{x^3}{3} - x]_1^3 = (9 - 3) - (\frac{1}{3} - 1) = 6 - (-\frac{2}{3}) = \frac{20}{3}$.
Summing these values: $I = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}$.

(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(x-[x]) \, dx$.
Since $[x]$ is the greatest integer function,we split the integral at integer points between $-\frac{\pi}{2} \approx -1.57$ and $\frac{\pi}{2} \approx 1.57$.
The integer points are $-1, 0, 1$.
$I = \int_{-\pi/2}^{-1} \sin(x-(-2)) \, dx + \int_{-1}^{0} \sin(x-(-1)) \, dx + \int_{0}^{1} \sin(x-0) \, dx + \int_{1}^{\pi/2} \sin(x-1) \, dx$.
$I = \int_{-\pi/2}^{-1} \sin(x+2) \, dx + \int_{-1}^{0} \sin(x+1) \, dx + \int_{0}^{1} \sin(x) \, dx + \int_{1}^{\pi/2} \sin(x-1) \, dx$.
Evaluating each integral:
$1$. $[-\cos(x+2)]_{-\pi/2}^{-1} = -\cos(1) + \cos(2-\pi/2) = -\cos 1 + \sin 2$.
$2$. $[-\cos(x+1)]_{-1}^{0} = -\cos 1 + \cos 0 = 1 - \cos 1$.
$3$. $[-\cos x]_{0}^{1} = -\cos 1 + \cos 0 = 1 - \cos 1$.
$4$. $[-\cos(x-1)]_{1}^{\pi/2} = -\cos(\pi/2-1) + \cos 0 = -\sin 1 + 1$.
Summing these: $I = (\sin 2 - \cos 1) + (1 - \cos 1) + (1 - \cos 1) + (1 - \sin 1) = 3 - 3\cos 1 + \sin 2 - \sin 1$.

(A) We need to evaluate $I = \int_{-1}^1 (f(x) - g(x)) \, dx = \int_{-1}^1 f(x) \, dx - \int_{-1}^1 g(x) \, dx$.
First,consider $f(x) = \operatorname{Max}\{x^3-4, x^4-4\}$. Since $x^4 \ge x^3$ for $x \in [-1, 0]$ and $x^4 \le x^3$ for $x \in [0, 1]$,we have $f(x) = x^4-4$ for $x \in [-1, 0]$ and $f(x) = x^3-4$ for $x \in [0, 1]$.
$\int_{-1}^1 f(x) \, dx = \int_{-1}^0 (x^4-4) \, dx + \int_0^1 (x^3-4) \, dx = [\frac{x^5}{5} - 4x]_{-1}^0 + [\frac{x^4}{4} - 4x]_0^1 = (0 - (-\frac{1}{5} + 4)) + (\frac{1}{4} - 4) = -\frac{19}{5} - \frac{15}{4} = -\frac{76+75}{20} = -\frac{151}{20}$.
Next,consider $g(x) = \operatorname{Min}\{x^2, x^3\}$. For $x \in [-1, 0]$,$x^3 \le x^2$,so $g(x) = x^3$. For $x \in [0, 1]$,$x^2 \le x^3$,so $g(x) = x^2$.
$\int_{-1}^1 g(x) \, dx = \int_{-1}^0 x^3 \, dx + \int_0^1 x^2 \, dx = [\frac{x^4}{4}]_{-1}^0 + [\frac{x^3}{3}]_0^1 = (0 - \frac{1}{4}) + (\frac{1}{3} - 0) = -\frac{1}{4} + \frac{1}{3} = \frac{1}{12}$.
Finally,$I = -\frac{151}{20} - \frac{1}{12} = \frac{-453 - 5}{60} = -\frac{458}{60} = -\frac{229}{30}$.
Wait,re-evaluating the options,let's re-check the integral calculation.
$\int_{-1}^1 f(x) dx = -\frac{151}{20}$. $\int_{-1}^1 g(x) dx = \frac{1}{12}$. The result is $-\frac{229}{30}$.
Given the options provided,there might be a typo in the question or options. However,based on the standard calculation,the result is $-\frac{229}{30}$.

(A) To evaluate the integral $I = \int_0^1 \frac{x^4+1}{x^6+1} dx$,we can factor the denominator as $x^6+1 = (x^2+1)(x^4-x^2+1)$.
However,a more effective method is to divide the numerator and denominator by $x^2$ or use partial fractions.
Alternatively,we can write $x^4+1 = (x^4-x^2+1) + x^2$.
Then $I = \int_0^1 \frac{x^4-x^2+1}{x^6+1} dx + \int_0^1 \frac{x^2}{x^6+1} dx$.
$I = \int_0^1 \frac{1}{x^2+1} dx + \int_0^1 \frac{x^2}{(x^3)^2+1} dx$.
For the first part,$\int_0^1 \frac{1}{x^2+1} dx = [\tan^{-1}(x)]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4}$.
For the second part,let $u = x^3$,then $du = 3x^2 dx$,so $x^2 dx = \frac{du}{3}$.
When $x=0, u=0$ and when $x=1, u=1$.
So,$\int_0^1 \frac{x^2}{(x^3)^2+1} dx = \frac{1}{3} \int_0^1 \frac{1}{u^2+1} du = \frac{1}{3} [\tan^{-1}(u)]_0^1 = \frac{1}{3} (\frac{\pi}{4} - 0) = \frac{\pi}{12}$.
Adding both parts,$I = \frac{\pi}{4} + \frac{\pi}{12} = \frac{3\pi + \pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.

(C) The integral represents the area under the curve $y = |2-x|$ from $x = -3$ to $x = 3$.
We can split the integral at the point where the expression inside the modulus is zero,which is $x = 2$.
$\int_{-3}^3 |2-x| dx = \int_{-3}^2 (2-x) dx + \int_{2}^3 (x-2) dx$
Evaluating the first part: $\int_{-3}^2 (2-x) dx = [2x - \frac{x^2}{2}]_{-3}^2 = (4 - 2) - (-6 - \frac{9}{2}) = 2 - (-10.5) = 12.5$.
Evaluating the second part: $\int_{2}^3 (x-2) dx = [\frac{x^2}{2} - 2x]_{2}^3 = (4.5 - 6) - (2 - 4) = -1.5 - (-2) = 0.5$.
Adding both parts: $12.5 + 0.5 = 13$.
Alternatively,using the geometric interpretation from the graph,the area consists of two right-angled triangles:
Triangle $1$ (base from $-3$ to $2$,height at $x=-3$ is $|2-(-3)|=5$): Area $= \frac{1}{2} \times 5 \times 5 = 12.5$.
Triangle $2$ (base from $2$ to $3$,height at $x=3$ is $|2-3|=1$): Area $= \frac{1}{2} \times 1 \times 1 = 0.5$.
Total Area $= 12.5 + 0.5 = 13$.

(D) Let $I = \int_0^5 [x] \, dx$.
Since $[x]$ is the greatest integer function,it changes its value at every integer point.
We can split the integral as:
$I = \int_0^1 [x] \, dx + \int_1^2 [x] \, dx + \int_2^3 [x] \, dx + \int_3^4 [x] \, dx + \int_4^5 [x] \, dx$
$I = \int_0^1 0 \, dx + \int_1^2 1 \, dx + \int_2^3 2 \, dx + \int_3^4 3 \, dx + \int_4^5 4 \, dx$
$I = 0(1-0) + 1(2-1) + 2(3-2) + 3(4-3) + 4(5-4)$
$I = 0 + 1 + 2 + 3 + 4 = 10$.

(C) We need to evaluate the integral $I = \int_{1}^{5} (|x-3| + |1-x|) dx$.
Since $x$ ranges from $1$ to $5$,we have $|1-x| = x-1$ because $x \ge 1$.
Now,we split the integral at $x=3$ due to the modulus $|x-3|$:
For $1 \le x < 3$,$|x-3| = 3-x$.
For $3 \le x \le 5$,$|x-3| = x-3$.
Thus,$I = \int_{1}^{3} (3-x + x-1) dx + \int_{3}^{5} (x-3 + x-1) dx$.
$I = \int_{1}^{3} 2 dx + \int_{3}^{5} (2x-4) dx$.
$I = [2x]_{1}^{3} + [x^2-4x]_{3}^{5}$.
$I = (6-2) + ((25-20) - (9-12))$.
$I = 4 + (5 - (-3)) = 4 + 8 = 12$.

(B) Let $I = \int_0^1 (\sqrt{10})^{2x} dx$.
Since $(\sqrt{10})^{2x} = (10^{1/2})^{2x} = 10^x$,the integral becomes:
$I = \int_0^1 10^x dx$.
Using the standard integral formula $\int a^x dx = \frac{a^x}{\ln a} + C$,we get:
$I = \left[ \frac{10^x}{\ln 10} \right]_0^1$.
Evaluating at the limits:
$I = \frac{10^1}{\ln 10} - \frac{10^0}{\ln 10} = \frac{10}{\ln 10} - \frac{1}{\ln 10} = \frac{9}{\ln 10}$.

(A) To evaluate the integral $\int_0^2 f(x) \, dx$,we split the interval $[0, 2]$ at $x = 1$ based on the definition of $f(x)$.
$\int_0^2 f(x) \, dx = \int_0^1 x^2 \, dx + \int_1^2 \sqrt{x} \, dx$
Evaluating the first part: $\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$
Evaluating the second part: $\int_1^2 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_1^2 = \left[ \frac{2}{3} x^{3/2} \right]_1^2 = \frac{2}{3} (2^{3/2} - 1^{3/2}) = \frac{2}{3} (2\sqrt{2} - 1)$
Adding both parts: $\frac{1}{3} + \frac{2}{3}(2\sqrt{2} - 1) = \frac{1 + 4\sqrt{2} - 2}{3} = \frac{4\sqrt{2} - 1}{3}$

(D) Let $I = \int_0^\pi \frac{\cos x}{\sqrt{1-\sin^2 x}} dx$.
Since $\sqrt{1-\sin^2 x} = |\cos x|$,the integral becomes $I = \int_0^\pi \frac{\cos x}{|\cos x|} dx$.
We know that $\cos x > 0$ for $x \in [0, \pi/2)$ and $\cos x < 0$ for $x \in (\pi/2, \pi]$.
Thus,we split the integral at $x = \pi/2$:
$I = \int_0^{\pi/2} \frac{\cos x}{\cos x} dx + \int_{\pi/2}^\pi \frac{\cos x}{-\cos x} dx$.
$I = \int_0^{\pi/2} 1 dx - \int_{\pi/2}^\pi 1 dx$.
$I = [x]_0^{\pi/2} - [x]_{\pi/2}^\pi$.
$I = (\pi/2 - 0) - (\pi - \pi/2) = \pi/2 - \pi/2 = 0$.

(A) Let $f(x) = \sqrt{3+x+x^2}$. Since $f(x)$ is an increasing function on $[1, 3]$,we have $f(1) \le f(x) \le f(3)$ for all $x \in [1, 3]$.
$f(1) = \sqrt{3+1+1} = \sqrt{5}$.
$f(3) = \sqrt{3+3+9} = \sqrt{15}$.
By the property of definite integrals,$\int_a^b f(x) dx \le (b-a) \times \max(f(x))$ and $\int_a^b f(x) dx \ge (b-a) \times \min(f(x))$.
Here,$a=1, b=3$,so $b-a = 2$.
Thus,$2 \times \sqrt{5} \le I \le 2 \times \sqrt{15}$.
Therefore,$I \in [2 \sqrt{5}, 2 \sqrt{15}]$.
Comparing this with the given options,the correct interval is $(2 \sqrt{5}, 2 \sqrt{15})$.

(A) Given,$\int_n^{n+1} g(x) dx = n^2, \forall n \in Z$.
We need to evaluate $\int_{-3}^3 g(x) dx$.
Using the additive property of definite integrals,we can write:
$\int_{-3}^3 g(x) dx = \int_{-3}^{-2} g(x) dx + \int_{-2}^{-1} g(x) dx + \int_{-1}^0 g(x) dx + \int_0^1 g(x) dx + \int_1^2 g(x) dx + \int_2^3 g(x) dx$.
Substituting the given values for each interval:
For $n = -3: \int_{-3}^{-2} g(x) dx = (-3)^2 = 9$.
For $n = -2: \int_{-2}^{-1} g(x) dx = (-2)^2 = 4$.
For $n = -1: \int_{-1}^0 g(x) dx = (-1)^2 = 1$.
For $n = 0: \int_0^1 g(x) dx = (0)^2 = 0$.
For $n = 1: \int_1^2 g(x) dx = (1)^2 = 1$.
For $n = 2: \int_2^3 g(x) dx = (2)^2 = 4$.
Summing these values:
$\int_{-3}^3 g(x) dx = 9 + 4 + 1 + 0 + 1 + 4 = 19$.

(D) We evaluate the integral $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx$ by splitting the interval based on the values where $[x]$ changes:
$\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx = \int_{\sqrt{3}}^{2}[x] \, dx + \int_{2}^{3}[x] \, dx + \int_{3}^{4}[x] \, dx + \int_{4}^{\sqrt{18}}[x] \, dx$
Since $\sqrt{3} \approx 1.732$ and $\sqrt{18} = 3\sqrt{2} \approx 4.242$,we have:
$\int_{\sqrt{3}}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{4} 3 \, dx + \int_{4}^{3\sqrt{2}} 4 \, dx$
$= (2 - \sqrt{3}) + 2(3 - 2) + 3(4 - 3) + 4(3\sqrt{2} - 4)$
$= 2 - \sqrt{3} + 2 + 3 + 12\sqrt{2} - 16$
$= (2 + 2 + 3 - 16) + 12\sqrt{2} - \sqrt{3}$
$= -9 + 12\sqrt{2} - \sqrt{3}$
Comparing this with $a + b\sqrt{2} + c\sqrt{3}$,we get $a = -9$,$b = 12$,and $c = -1$.
Therefore,$a + b + c = -9 + 12 - 1 = 2$.

(C) Let $I = \int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$.
Using the identity $\cos^2 A - \cos^2 B = \sin(B-A) \sin(B+A)$,we have:
$A = \frac{3\pi}{8} - \frac{x}{4}$ and $B = \frac{11\pi}{8} + \frac{x}{4}$.
$B - A = \left(\frac{11\pi}{8} + \frac{x}{4}\right) - \left(\frac{3\pi}{8} - \frac{x}{4}\right) = \frac{8\pi}{8} + \frac{2x}{4} = \pi + \frac{x}{2}$.
$B + A = \left(\frac{11\pi}{8} + \frac{x}{4}\right) + \left(\frac{3\pi}{8} - \frac{x}{4}\right) = \frac{14\pi}{8} = \frac{7\pi}{4}$.
Thus,the integrand becomes $\sin(\pi + \frac{x}{2}) \sin(\frac{7\pi}{4}) = (-\sin \frac{x}{2}) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} \sin \frac{x}{2}$.
$I = \int_0^\pi \frac{1}{\sqrt{2}} \sin \frac{x}{2} dx = \frac{1}{\sqrt{2}} \left[ -2 \cos \frac{x}{2} \right]_0^\pi$.
$I = \frac{-2}{\sqrt{2}} (\cos \frac{\pi}{2} - \cos 0) = -\sqrt{2} (0 - 1) = \sqrt{2}$.

(C) Let $I = \int_0^{\pi^2 / 4} (2 \sin \sqrt{x} + \sqrt{x} \cos \sqrt{x}) \, dx$.
Substitute $t = \sqrt{x}$,so $x = t^2$ and $dx = 2t \, dt$.
When $x = 0$,$t = 0$. When $x = \frac{\pi^2}{4}$,$t = \frac{\pi}{2}$.
Substituting these into the integral:
$I = \int_0^{\pi/2} (2 \sin t + t \cos t) (2t) \, dt = \int_0^{\pi/2} (4t \sin t + 2t^2 \cos t) \, dt$.
Using integration by parts on the second term $\int 2t^2 \cos t \, dt$:
Let $u = 2t^2$ and $dv = \cos t \, dt$,then $du = 4t \, dt$ and $v = \sin t$.
$\int 2t^2 \cos t \, dt = 2t^2 \sin t - \int 4t \sin t \, dt$.
Substituting this back into the expression for $I$:
$I = \int_0^{\pi/2} 4t \sin t \, dt + [2t^2 \sin t]_0^{\pi/2} - \int_0^{\pi/2} 4t \sin t \, dt$.
$I = [2t^2 \sin t]_0^{\pi/2} = 2(\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) - 0 = 2(\frac{\pi^2}{4})(1) = \frac{\pi^2}{2}$.

(C) Let $I = \int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos^8 x} \, dx = \int_{-\pi / 4}^{\pi / 4} (\sec^2 x)^3 \sec^2 x \, dx$.
Since the integrand is an even function,$I = 2 \int_{0}^{\pi / 4} (1 + \tan^2 x)^3 \sec^2 x \, dx$.
Let $\tan x = t$,then $\sec^2 x \, dx = dt$.
When $x = 0, t = 0$ and when $x = \pi / 4, t = 1$.
$I = 2 \int_{0}^{1} (1 + t^2)^3 \, dt = 2 \int_{0}^{1} (1 + 3t^2 + 3t^4 + t^6) \, dt$.
$I = 2 \left[ t + t^3 + \frac{3t^5}{5} + \frac{t^7}{7} \right]_{0}^{1} = 2 \left( 1 + 1 + \frac{3}{5} + \frac{1}{7} \right)$.
$I = 2 \left( 2 + \frac{21 + 5}{35} \right) = 2 \left( 2 + \frac{26}{35} \right) = 2 \left( \frac{70 + 26}{35} \right) = 2 \left( \frac{96}{35} \right) = \frac{192}{35}$.

(D) We know that for $y > 0$,$\tan ^{-1}(y) + \tan ^{-1}(1/y) = \frac{\pi}{2}$.
Let $y = \frac{x}{x^2+1}$. Since $x \in [1, 2]$,$y > 0$.
Therefore,the integrand simplifies as follows:
$\tan ^{-1}\left(\frac{x}{x^2+1}\right) + \tan ^{-1}\left(\frac{x^2+1}{x}\right) = \frac{\pi}{2}$.
Now,evaluate the integral:
$I = \int_1^2 \frac{\pi}{2} d x$
$I = \frac{\pi}{2} [x]_1^2$
$I = \frac{\pi}{2} (2 - 1) = \frac{\pi}{2}$.

(A) To evaluate the integral $\int_{0}^{2} x e^{x} dx$,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: $\int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x}$.
Now,applying the limits from $0$ to $2$:
$\left[ x e^{x} - e^{x} \right]_{0}^{2} = (2 e^{2} - e^{2}) - (0 \cdot e^{0} - e^{0})$.
$= (e^{2}) - (0 - 1) = e^{2} + 1$.

(C) We are given the integral $I = \int_{-1}^1 \frac{|x|}{x} \, dx$.
First,we define the function $f(x) = \frac{|x|}{x}$.
Based on the definition of the absolute value function,we have:
$f(x) = \begin{cases} -\frac{x}{x} = -1, & x < 0 \\ \frac{x}{x} = 1, & x > 0 \end{cases}$
Since the function is discontinuous at $x = 0$,we split the integral at $x = 0$:
$I = \int_{-1}^0 (-1) \, dx + \int_0^1 (1) \, dx$
Evaluating the integrals:
$I = [-x]_{-1}^0 + [x]_0^1$
$I = (-(0) - (-(-1))) + (1 - 0)$
$I = (-1) + 1 = 0$
Thus,the value of the integral is $0$.

(A) Given,$\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$.
First,evaluate the integral $\int_a^b x^3 dx = 0$:
$\left[\frac{x^4}{4}\right]_a^b = 0 \Rightarrow \frac{b^4 - a^4}{4} = 0 \Rightarrow b^4 = a^4$.
This implies $b = a$ or $b = -a$. Since $a$ and $b$ are limits of integration,we assume $a \neq b$,so $b = -a$.
Next,evaluate the integral $\int_a^b x^2 dx = \frac{2}{3}$:
$\left[\frac{x^3}{3}\right]_a^b = \frac{2}{3} \Rightarrow \frac{b^3 - a^3}{3} = \frac{2}{3} \Rightarrow b^3 - a^3 = 2$.
Substitute $b = -a$ into the equation:
$(-a)^3 - a^3 = 2 \Rightarrow -a^3 - a^3 = 2 \Rightarrow -2a^3 = 2$.
$a^3 = -1 \Rightarrow a = -1$.
Since $b = -a$,we have $b = -(-1) = 1$.
Thus,$a = -1$ and $b = 1$.

(A) Let $I = \int_0^{\pi / 4} \frac{d x}{\cos ^3 x \sqrt{2 \sin 2 x}}$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int_0^{\pi / 4} \frac{d x}{\cos ^3 x \sqrt{4 \sin x \cos x}} = \int_0^{\pi / 4} \frac{d x}{2 \cos ^4 x \sqrt{\tan x}}$.
$I = \frac{1}{2} \int_0^{\pi / 4} \frac{\sec^2 x (1 + \tan^2 x)}{\sqrt{\tan x}} d x$.
Let $\tan x = t^2$,then $\sec^2 x d x = 2t dt$.
When $x = 0, t = 0$ and when $x = \frac{\pi}{4}, t = 1$.
Substituting these into the integral:
$I = \frac{1}{2} \int_0^1 \frac{(1 + t^4) \cdot 2t dt}{t} = \int_0^1 (1 + t^4) dt$.
$I = [t + \frac{t^5}{5}]_0^1 = 1 + \frac{1}{5} = \frac{6}{5}$.

(C) We know that the absolute value function $|x|$ is defined as:
$|x| = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases}$
Therefore,the integral can be split at $x = 0$:
$I = \int_{-1}^2 |x| \, dx = \int_{-1}^0 (-x) \, dx + \int_0^2 (x) \, dx$
Evaluating the first part:
$\int_{-1}^0 (-x) \, dx = \left[ -\frac{x^2}{2} \right]_{-1}^0 = 0 - \left( -\frac{(-1)^2}{2} \right) = 0 - (-\frac{1}{2}) = \frac{1}{2}$
Evaluating the second part:
$\int_0^2 (x) \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - 0 = \frac{4}{2} = 2$
Adding both parts:
$I = \frac{1}{2} + 2 = \frac{1+4}{2} = \frac{5}{2}$

(C) Let $I = \int_0^3 |x^2 - 3x + 2| dx$.
Since $x^2 - 3x + 2 = (x - 1)(x - 2)$,the expression inside the modulus changes sign at $x = 1$ and $x = 2$.
For $x \in [0, 1]$,$x^2 - 3x + 2 \ge 0$.
For $x \in (1, 2)$,$x^2 - 3x + 2 < 0$.
For $x \in [2, 3]$,$x^2 - 3x + 2 \ge 0$.
Thus,$I = \int_0^1 (x^2 - 3x + 2) dx - \int_1^2 (x^2 - 3x + 2) dx + \int_2^3 (x^2 - 3x + 2) dx$.
Evaluating the integral $\int (x^2 - 3x + 2) dx = \frac{x^3}{3} - \frac{3x^2}{2} + 2x + C$.
$I = [\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_0^1 - [\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_1^2 + [\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_2^3$.
$I = (\frac{1}{3} - \frac{3}{2} + 2) - [(\frac{8}{3} - 6 + 4) - (\frac{1}{3} - \frac{3}{2} + 2)] + [(\frac{27}{3} - \frac{27}{2} + 6) - (\frac{8}{3} - 6 + 4)]$.
$I = (\frac{2 - 9 + 12}{6}) - [(\frac{2}{3}) - (\frac{5}{6})] + [(\frac{3}{2}) - (\frac{2}{3})]$.
$I = \frac{5}{6} - [-\frac{1}{6}] + \frac{5}{6} = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}$.

(D) We have,$\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$.
Integrating both sides,we get $[\tan^{-1} x]_0^b = [\tan^{-1} x]_b^{\infty}$.
Substituting the limits,$\tan^{-1}(b) - \tan^{-1}(0) = \tan^{-1}(\infty) - \tan^{-1}(b)$.
Since $\tan^{-1}(0) = 0$ and $\tan^{-1}(\infty) = \frac{\pi}{2}$,we have $\tan^{-1}(b) = \frac{\pi}{2} - \tan^{-1}(b)$.
Adding $\tan^{-1}(b)$ to both sides,we get $2 \tan^{-1}(b) = \frac{\pi}{2}$.
Therefore,$\tan^{-1}(b) = \frac{\pi}{4}$.
Taking tangent on both sides,$b = \tan(\frac{\pi}{4}) = 1$.

(C) Let $I = \int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$.
Since $\cosh x = \frac{e^x + e^{-x}}{2}$,we substitute this into the integral:
$I = \int_{-1}^1 \frac{e^x + e^{-x}}{2(1 + e^{2x})} d x$.
Factor out $e^x$ from the numerator term $e^x + e^{-x}$:
$e^x + e^{-x} = e^x(1 + e^{-2x}) = e^x \left(1 + \frac{1}{e^{2x}}\right) = e^x \left(\frac{e^{2x} + 1}{e^{2x}}\right) = \frac{1 + e^{2x}}{e^x}$.
Substituting this back into the integral:
$I = \frac{1}{2} \int_{-1}^1 \frac{1 + e^{2x}}{e^x(1 + e^{2x})} d x$.
Canceling the term $(1 + e^{2x})$:
$I = \frac{1}{2} \int_{-1}^1 e^{-x} d x$.
Evaluating the integral:
$I = \frac{1}{2} [-e^{-x}]_{-1}^1 = -\frac{1}{2} [e^{-1} - e^1] = \frac{1}{2} (e^1 - e^{-1}) = \frac{e^2 - 1}{2e}$.

(B) We have,
$\int_2^3 \frac{d x}{x^2-x} = \int_2^3 \frac{1}{x(x-1)} d x$
Using partial fractions,$\frac{1}{x(x-1)} = \frac{1}{x-1} - \frac{1}{x}$
So,$\int_2^3 \left( \frac{1}{x-1} - \frac{1}{x} \right) d x = [\log |x-1| - \log |x|]_2^3$
$= [\log |\frac{x-1}{x}|]_2^3$
$= \log |\frac{3-1}{3}| - \log |\frac{2-1}{2}|$
$= \log \frac{2}{3} - \log \frac{1}{2}$
$= \log \left( \frac{2/3}{1/2} \right) = \log \frac{4}{3}$

(B) We split the integral based on the definition of the greatest integer function $[x]$:
$\int_{-0.5}^{1.5} x^2[x] d x = \int_{-0.5}^{0} x^2[x] d x + \int_{0}^{1} x^2[x] d x + \int_{1}^{1.5} x^2[x] d x$
For $-0.5 \le x < 0$,$[x] = -1$.
For $0 \le x < 1$,$[x] = 0$.
For $1 \le x < 1.5$,$[x] = 1$.
Substituting these values:
$\int_{-0.5}^{0} x^2(-1) d x + \int_{0}^{1} x^2(0) d x + \int_{1}^{1.5} x^2(1) d x$
$= -\int_{-0.5}^{0} x^2 d x + 0 + \int_{1}^{1.5} x^2 d x$
$= -\left[ \frac{x^3}{3} \right]_{-0.5}^{0} + \left[ \frac{x^3}{3} \right]_{1}^{1.5}$
$= -\left( 0 - \frac{(-0.5)^3}{3} \right) + \left( \frac{(1.5)^3}{3} - \frac{1^3}{3} \right)$
$= -\left( 0 - \frac{-0.125}{3} \right) + \left( \frac{3.375}{3} - \frac{1}{3} \right)$
$= -\frac{0.125}{3} + \frac{2.375}{3} = \frac{2.25}{3} = 0.75 = \frac{3}{4}$

(B) Given the integral $\int_0^a 2^{[x]} dx = 127$,where $a \in Z^{+}$.
We can split the integral into unit intervals:
$\int_0^1 2^{[x]} dx + \int_1^2 2^{[x]} dx + \dots + \int_{a-1}^a 2^{[x]} dx = 127$
Since $[x] = k$ for $x \in [k, k+1)$,the integral becomes:
$\int_0^1 2^0 dx + \int_1^2 2^1 dx + \int_2^3 2^2 dx + \dots + \int_{a-1}^a 2^{a-1} dx = 127$
Evaluating each term:
$2^0(1-0) + 2^1(2-1) + 2^2(3-2) + \dots + 2^{a-1}(a-(a-1)) = 127$
$2^0 + 2^1 + 2^2 + \dots + 2^{a-1} = 127$
This is a geometric progression with $n = a$ terms,first term $1$,and common ratio $2$:
$\frac{1(2^a - 1)}{2 - 1} = 127$
$2^a - 1 = 127$
$2^a = 128$
$2^a = 2^7$
Therefore,$a = 7$.

(C) The given integral is $I = \int_{\frac{3 \pi}{4}}^\pi \left[ \sin x + \left[ \frac{4 x}{\pi} \right] \right] dx$.
Since $[a + n] = [a] + n$ for any integer $n$,we have $\left[ \sin x + \left[ \frac{4 x}{\pi} \right] \right] = [\sin x] + \left[ \frac{4 x}{\pi} \right]$.
For the interval $\frac{3 \pi}{4} \leq x \leq \pi$,the value of $\sin x$ lies between $0$ and $\frac{1}{\sqrt{2}}$.
Since $0 \leq \sin x < 1$,the greatest integer function $[\sin x] = 0$.
Now,for $\frac{3 \pi}{4} \leq x < \pi$,we have $3 \leq \frac{4 x}{\pi} < 4$.
Therefore,$\left[ \frac{4 x}{\pi} \right] = 3$.
Substituting these values into the integral,we get $I = \int_{\frac{3 \pi}{4}}^\pi (0 + 3) dx$.
$I = 3 \int_{\frac{3 \pi}{4}}^\pi dx = 3 [x]_{\frac{3 \pi}{4}}^\pi = 3 \left( \pi - \frac{3 \pi}{4} \right) = 3 \left( \frac{\pi}{4} \right) = \frac{3 \pi}{4}$.

(B) Given $f(x) = \sin(\tan^{-1} x)$.
Using the identity $\sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}}$,we have $f(x) = \frac{x}{\sqrt{1+x^2}}$.
We need to evaluate $I = \int_0^1 x f''(x) dx$.
Using integration by parts,let $u = x$ and $dv = f''(x) dx$. Then $du = dx$ and $v = f'(x)$.
$I = [x f'(x)]_0^1 - \int_0^1 f'(x) dx$.
$I = (1 \cdot f'(1) - 0 \cdot f'(0)) - [f(1) - f(0)]$.
First,calculate $f'(x) = \frac{d}{dx} \left( \frac{x}{\sqrt{1+x^2}} \right) = \frac{\sqrt{1+x^2} - x \cdot \frac{x}{\sqrt{1+x^2}}}{1+x^2} = \frac{1+x^2-x^2}{(1+x^2)^{3/2}} = \frac{1}{(1+x^2)^{3/2}}$.
Now,$f'(1) = \frac{1}{(1+1^2)^{3/2}} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}}$.
Also,$f(1) = \sin(\tan^{-1} 1) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $f(0) = \sin(\tan^{-1} 0) = 0$.
Substituting these values into the expression for $I$:
$I = \frac{1}{2\sqrt{2}} - (\frac{1}{\sqrt{2}} - 0) = \frac{1}{2\sqrt{2}} - \frac{2}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}}$.