Definite Integration

$ \begin{aligned} & I=\int_0^1 x \sin ^{-1} x \cdot d x \\ & \because \int u v \cdot d x=u \int v d x-\int\left(u^{\prime} \int u d x\right) d x \end{aligned} $

Here, $u=\sin ^{-1} x, v=x$

$ \begin{aligned} &\because I=\left[\sin ^{-1} x \int x \cdot d x-\int\left(\frac{1}{\sqrt{1-x^2}} \int x \cdot d x\right) d x\right]_0^1 \\ & =\left[\sin ^{-1} x \cdot \frac{x^2}{2}-\int \frac{x^2}{2 \sqrt{1-x^2}} d x\right]_0^1 \\ & =\left(\sin ^{-1}(1) \cdot \frac{1}{2}-\sin ^{-1}(0) \cdot 0\right)-\int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} d x \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} d x \end{aligned} $

Let $x=\sin \theta \Rightarrow d x=\cos \theta \cdot d \theta$ at $x=1 \Rightarrow \theta=\pi / 2$ and $x=0 \Rightarrow \theta=0$

$ \begin{aligned} & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2} \frac{\sin ^2 \theta \cdot \cos \theta}{\cos \theta} d \theta \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2} \sin ^2 \theta \cdot d \theta \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2}\left(\frac{1}{2}-\frac{\cos 2 \theta}{2}\right) d \theta \\ & =\frac{\pi}{4}-\frac{1}{2}\left[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right]_0^{\pi / 2} \\ & =\frac{\pi}{4}-\frac{1}{2}\left[\frac{\pi}{4}-0-0\right]=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8} \end{aligned} $

$\because I=\int_{-\pi / 2}^{\pi / 2} \sin (x-[x]) d x$

for $x \in[-\pi / 2,-1] \Rightarrow[x]=-2$

and $x-[x]=x+2$

for $x \in[-1,0) \Rightarrow[x]=-1$

and $x-[x]=x+1$

for $x \in[0,1] \Rightarrow[x]=0$ and $x-[x]=x$

for $x \in\left[1, \frac{\pi}{2}\right] \Rightarrow x-[x]=x-1$

$ \begin{aligned} \therefore I=\int_{-\pi / 2}^{-1} \sin (x & +2 d x+\int_{-1}^0 \sin (x+1) d x +\int_0^1 \sin x \cdot d x+\int_1^{\pi / 2} \sin (x-1) d x \end{aligned} $

$ \begin{aligned} = & {[-\cos (x+2)]_{-\pi / 2}^1+[-\cos (x+1)]_{-1}^0 } +[-\cos x]_0^1+[-\cos (x-1)]_1^{\pi / 2} \\ = & -\cos (1)+\cos (2-\pi / 2)-\cos (1)+\cos (0) -\cos (1)+\cos (0)-\cos (\pi / 2-1)+\cos (0) \\ = & -3 \cos (1)+3+\cos (-(\pi / 2-2)) -\cos (\pi / 2-1) \\ = & 3-3 \cos (1)+\sin (2)-\sin (1) \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \int_0^2 x^2(2-x)^5 d x \\ & =\int_0^2 x^2\left(32-80 x+80 x^2-40 x^3\right. \left.+10 x^4-x^5\right) d x \\ & =\left[\begin{array}{rl} \frac{32 x^3}{3}-\frac{80 x^4}{4}+\frac{80 x^5}{5}-\frac{40 x^6}{6} \left.+\frac{10 x^7}{7}-\frac{x^8}{8}\right]_0^2 \end{array}\right. \\ & =\left(\frac{32 \times 8}{3}-20 \times 16+16 \times 32-\frac{20}{3} \times 64+\frac{10}{7}\right. \left.\times 128-\frac{256}{8}\right)-0\\ & =\frac{256}{3}+16 \times 12-\frac{1280}{3}+\frac{1280}{7}-32 \\ & =\frac{-1024}{3}+\frac{1280}{7}+160 \\ & =\frac{-7168+3840+3360}{21}=\frac{32}{21} \end{aligned} \end{aligned} $

$\because f(x)=\max \left\{x^3-4, x^4-4\right\}$,

$ \begin{aligned} & g(x)=\min \left\{x^2, x^3\right\} \\ & \text { at } x=-1 \Rightarrow x^2=1, x^3=-1 \\ & \Rightarrow \min \{1,-1\}=-1=(-1)^3 \end{aligned} $

Similarly, at $x=0.5, \Rightarrow x^2=0.25$,

$ \begin{aligned} & x^3=0.125 \\ & \Rightarrow \min \{0.25,0.125\}=0.125=(0.5)^3 \end{aligned} $

So, for $x \in[-1,1] \Rightarrow g(x)=x^3$

Similarly, analysis of $f(x)$

When, $x^3-4=x^4-4$

$ \Rightarrow \quad x^3(x-1)=0 $

So, $x=0,1$

for $x<0 \Rightarrow x^4-4>x^3-4$

for $0x^4-x$

thus, $f(x)=\left\{\begin{array}{lc}x^4-4, & -1 \leq x \leq 0 \\ x^3-4, & 0 < x \leq 1\end{array}\right.$

$ \begin{aligned} & \text { Therefore, } f(x)-g(x) \\ & =\left\{\begin{array}{cc} \left(x^4-4\right)-x^3=x^4-x^3-4, & -1 \leq x \leq 0 \\ \left(x^3-4\right)-x^3=-4, & 0 < x \leq 1 \end{array}\right. \\ & \therefore \int_{-1}^1(f(x)-g(x)) d x=\int_{-1}^0\left(x^4-x^3-4\right) d x +\int_0^1-4 \cdot d x \end{aligned} $

$ \begin{aligned} & =\left[\frac{x^5}{5}-\frac{x^4}{4}-4 x\right]_{-1}^0-4[x]_0^1 \\ & =\left[0-\left(-\frac{1}{5}-\frac{1}{4}+4\right)-4(1)-0\right] \end{aligned} $

$ =-8+\frac{1}{5}+\frac{1}{4}=\frac{-160+5+4}{20}=\frac{-151}{20} $

$ \begin{aligned} & \text { Let } I=\int_0^1 \frac{2 x+5}{x^2+3 x+2} d x \\ & \begin{aligned} \frac{2 x+5}{x^2+3 x+2} & =\frac{A}{(x+1)}+\frac{B}{(x+2)} \\ 2 x+5 & =A(x+2)+B(x+1) \\ 2 & =A+B, 5=2 A+B \end{aligned} \end{aligned} $

Solving we get $A=3, B=-1$

$ \begin{aligned} I & =\int_0^1\left(\frac{3}{x+1}-\frac{1}{x+2}\right) d x \\ & =3[\ln \mid(x+1)]_0^1-[\ln |x+2|]_0^1 \\ & =3 \ln 2-\ln 3+\ln 2 \\ & =4 \ln 2-\ln 3 \\ & =4 \log 2-\log 3=\log \left(\frac{2^4}{3}\right) \\ & =\log \left(\frac{16}{3}\right) \end{aligned} $

Let $I=\int_0^1 x^{\frac{5}{2}}(1-x)^{\frac{3}{2}} d x$

$ \text { } \begin{aligned}Put\,\,\,\, & x=\sin ^2 \theta \\ & d x=2 \sin \theta \cos \theta d \theta \\ \Rightarrow I= & \int_0^{\frac{\pi}{2}} \sin ^5 \theta \cdot\left(\cos ^2 \theta\right)^{\frac{3}{2}} \cdot 2 \sin \theta \cdot \cos \theta d \theta \\ = & 2 \int_0^{\frac{\pi}{2}} \sin ^6 \theta \cdot \cos ^4 \theta d \theta \end{aligned} $

Using Wali's theorem,

$ =2\left(\frac{5 \cdot 3 \cdot 1 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}\right)=\frac{3 \pi}{256} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \lim _{n \rightarrow-\infty} \frac{1}{n^2} \cdot \sec ^2 \frac{1}{n^2}+ & \frac{2}{n^2} \sec ^2 \frac{4}{n^2} +\ldots+\frac{1}{n^2} \sec ^2 1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\lim _{n \rightarrow-\infty} \sum_{r=1}^n \frac{r}{n^2} \cdot \sec ^2 \frac{r^2}{n^2} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{r}{n}\right) \sec ^2\left(\frac{r}{n}\right)^2 \\ & =\int_0^1 x \cdot \sec ^2 x^2 d x \\ & 1 \leq r \leq n \\ & \frac{1}{n} \leq \frac{r}{n} \leq 1 \\ \because \quad & n \rightarrow \infty \\ & 0 \leq \frac{r}{n}<1 \end{aligned}\\ &\text { Put } \quad x^2=t\\ &\begin{aligned} & \Rightarrow \quad 2 x d x=d t \\ & \int_0^1 \sec ^2 t \frac{d t}{2}=\frac{1}{2}[\tan t]_0^1 \\ & \Rightarrow \quad \frac{\tan 1}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^\pi\left(\sin ^5 x \cos ^3 x+\sin ^4 x\right. \\ & \begin{array}{r} \left.\cos ^4 x+\sin ^3 x \cos ^4 x\right) d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow I=\int_0^\pi\left[\sin ^5(\pi-x) \cos ^3(\pi-x)\right. \\ +\sin ^4(\pi-x) \cos ^4(\pi-x) \\ \left.+\sin ^3(\pi-x) \cdot \cos ^4(\pi-x)\right] d x \\ \Rightarrow I=\int_0^\pi-\sin ^5 x \cos ^3 x+\sin ^4 x \cos ^4 x \\ +\sin ^3 x \cos ^4 x d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 2 I=2 \int_0^\pi\left(\sin ^4 x \cos ^4 x+\sin ^3 x \cos ^4 x\right) d x \\ & I=\int_0^\pi \frac{\sin ^4 2 x}{16} d x+\int_0^\pi \sin ^3 x \cdot \cos ^4 x d x \end{aligned} $

$ \begin{aligned} & \cos x=t \\ & \Rightarrow-\sin x d x=d t \\ & =\frac{1}{16} \int_0^\pi \frac{(1-\cos 4 x)^2}{4} d x+\int_1^{-1} t^4 \cdot\left(1-t^2\right)(-d t) \\ & =\frac{1}{64} \int_0^\pi\left(1-2 \cos 4 x+\cos ^2 4 x\right) d x \text {. } +\int_1^{-1} t^4\left(t^2-1\right) d t \\ & =\frac{1}{64} \int_0^\pi\left(1-2 \cos 4 x+\frac{1+\cos 8 x}{2}\right) d x +\left(\frac{t^7}{5}-\frac{t^5}{5}\right)_1^{-1} \\ & =\frac{1}{64} \int_0^\pi\left(\frac{3}{2}-2 \cos 4 x+\frac{1}{2} \cos 8 x\right) d x +\left(\frac{2}{5}-\frac{2}{7}\right) \\ & =\frac{3}{128}[x]_0^\pi-\frac{2}{64}\left[\frac{\sin 4 x}{4}\right]_0^\pi+\frac{1}{128}\left[\frac{\sin 8 x}{8}\right]_0^\pi +\frac{4}{35} \\ & \Rightarrow \frac{3 \pi}{128}+\frac{4}{35} \end{aligned} $

$ \begin{aligned} & \text {} \begin{aligned} &I=\int_0^1 \frac{x^4+1}{x^6+1} d x \\ & \text { put } x^6+1=\left(x^2\right)^3+1 \\ &=\left(x^2+1\right)\left(x^4-x^2+1\right) \\ &=\int_0^1 \frac{\left(x^4+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)} d x \\ & \Rightarrow \frac{x^4+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)} \\ &=\frac{A}{x^2+1}+\frac{B x^2+C}{x^4-x^2+1} \\ & x^4+1=A\left(x^4-x^2+1\right) +B x^2\left(x^2+1\right)+C x^2+C \end{aligned} \end{aligned} $

$ \begin{array}{lll} \hline 1=A+B & -A+B+C=0 & A+C=1 \\ \hline B=1-A & A-B-C=0 & C=1-A \\ \hline \end{array} $

$ \begin{aligned} & \therefore \quad A-(1-A)-(1-A)=0 \\ & \Rightarrow \quad A-1+A-1+A=0 \\ & \Rightarrow \quad 3 A=2 \\ & \Rightarrow \quad A=2 / 3 \\ & B=C=1-2 / 3=1 / 3 \\ & I=\frac{2}{3} \int_0^1 \frac{1}{x^2+1} d x-\frac{1}{3} \int \frac{x^2+1}{x^4-x^2+1} d x \\ & =\frac{2}{3}\left[\tan ^{-1} x\right]_0^1-\frac{1}{3} \int_0^1 \frac{1+1 / x^2}{x^2+1 / x^2-1} d x \\ & =\frac{2}{3}\left(\frac{\pi}{4}\right)-\frac{1}{3} \int_0^1 \frac{\left(1+1 / x^2\right)}{\left(x-\frac{1}{x}\right)^2+1} d x \\ & =\frac{\pi}{6}-\frac{1}{3}\left[\tan ^{-1}(x-1 / x)\right]_0^1 \\ & =\frac{\pi}{6}-1 / 3\left(\frac{-\pi}{2}\right) \\ & \Rightarrow \frac{\pi}{6}+\pi / 6=\frac{\pi}{3} \end{aligned} $

$I=\int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x$

Let $u=2 x$, then $d u=2 d x$,

$ \text { so } d x=\frac{1}{2} d u $

When $x=-2 \pi, u=2(-2 \pi)=-4 \pi$

$ \begin{aligned} I & =\int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x \\ & =\frac{1}{2} \int_{-4 \pi}^{4 \pi} \sin ^4(u) \cos ^6(u) d u \end{aligned} $

Since, $\sin (u)$ and $\cos (u)$ are periodic with a period of $2 \pi$,

So, $\sin ^4(u)$ and $\cos ^6(u)$ are also periodic with a period of $2 \pi\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} & =\frac{1}{2} \cdot 2 \int_{-2 \pi}^{2 \pi} \sin ^4(u) \cos ^6(u) d u \\ & =\int_{-2 \pi}^{2 \pi} \sin ^4(u) \cos ^6(u) d u \\ & =\frac{1}{2} \cdot 4 \int_0^{2 \pi} \sin ^4(u) \cos ^6(u) d u \\ & =2 \int_0^{2 \pi} \sin ^4(u) \cos ^6(u) d u \end{aligned} $

Since, the function, $f(u)=\sin ^4(u) \cos ^6(u)$ is an even function.

$ \begin{aligned} & \therefore 2 \int_0^{2 \pi} \sin ^4(u) \cos ^6(u) d u \\ & \quad=2 \cdot 2 \int_0^\pi \sin ^4(u) \cos ^6(u) d u \\ & \quad=4 \int_0^\pi \sin ^4(u) \cos ^6(u) d u \\ & \quad=4 \cdot 2 \int_0^{\pi / 2} \sin ^4(u) \cos ^6(u) d u \\ & \quad=8 \int_0^{\pi / 2} \sin ^4(u) \cos ^6(u) d u \end{aligned} $

We know that, $\int_0^{\pi / 2} \sin ^m(u) \cos ^n(u) d u$

$ =\frac{(m-1)(m-3) \ldots(1 \text { or } 2)(n-1)(n-3)\ldots(1 \text { or } 2)}{(m+n)(m+n-2) \ldots(1 \text { or } 2)} \times \alpha $

where, $\alpha=\frac{\pi}{2}$ if both $m$ and $n$ are even, and $\alpha=1$ othernels.

Here, $m=4, n=6$ both are even.

So, $\int_0^{\pi / 2} \sin ^4(4) \cos ^6(4) d u$

$ \begin{aligned} & =\frac{(4-1)(4-3)(6-1)(6-3)(6-5)}{(4+6)(4+6-2)(4+6-4)} \cdot \frac{\pi}{2} \\ & \quad \begin{array}{c} (4+6-6)(4+6-8) \\ =\frac{3 \cdot 1 \cdot 5 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}=\frac{45}{3840} \cdot \frac{\pi}{2}=\frac{3}{256} \cdot \frac{\pi}{2} \\ =\frac{3 \pi}{512} \\ \therefore 8 \int_0^{\pi / 2} \sin ^4(u) \cos ^6(u) d u \\ \quad=8 \cdot \frac{3 \pi}{512}=\frac{24 \pi}{512}=\frac{3 \pi}{64} \\ \therefore 8 \int_0^{\frac{\pi}{2}} \sin ^4(u) \cos ^6(u) d u=\frac{3 \pi}{64} \end{array} \end{aligned} $

$ \begin{aligned} & \text { Given, } f(t)=\int_0^t \tan ^{(2 n-1)} x d x, n \in \\ & \text { So, } f(t+\pi)=\int_0^{t+\pi} \tan ^{(2 n-1)} x d x \\ & =\int_0^t \tan ^{(2 n-1)} x d x+\int_t^{t+\pi} \tan ^{(2 n-1)} x d x \\ & \text { Let } u=x-\pi \\ & \Rightarrow d u=d x \end{aligned} $

Limit, when $x=t, u=t-\pi$ and

$ x=t+\pi, u=t $

So, $f(t+\pi)=f(t)+\int_t^t \tan ^{(2 n-1)}(u+\pi) d u$

$ \begin{aligned} & \text { Since, } \tan (u+\pi)=\tan u \\ & =f(t)+\int_{t-\pi}^t \tan ^{(2 n-1}(u) d u \\ & =f(t)+\int_{t-\pi}^0 \tan ^{(2 n-1)}(u) d u +\int_0^t \tan ^{(2 n-1)}(u) d u \end{aligned} $

Let $v=-u$, the $d v=-d u$

When $u=t-\pi, v=\pi-t$, when $u=0$, $v=0$, so we get

$ \begin{aligned} =f(t)-\int_{\pi-t}^0 \tan ^{(2 n-1)}(-v) d v & +\int_0^t \tan ^{(2 n-1)}(u) d u \\ =f(t)+\int_{\pi-t}^0 \tan ^{(2 n-1)}(v) d v & +\int_0^t \tan ^{(2 n-1)}(u) d u \end{aligned} $

$ \begin{aligned} & =f(t)+\int_{\pi-t}^0 \tan ^{(2 n-1)}(v) d v+f(t) \\ & =2 f(t)-\int_0^{\pi-t} \tan ^{(2 n-1)}(v) d v \\ & =2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v-\int_{\pi-t}^\pi \tan ^{(2 n-1)}(v) d v \\ & \text { Let } w=v-\pi, \text { then } d w=d v \\ & \text { When } v=\pi-t, w=-t, \text { when } v=\pi \\ & w=0 \\ & =2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v -\int_{-t}^0 \tan ^{(2 n-1)}(w+\pi) d w \\ & 2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v -\int_{-t}^0 \tan ^{(2 n-1)}(w) d w \end{aligned} $

Let $z=-w$, then $d z=-d w$ when $w=-t$, $z=t$ and $w=0, z=0$

$ \begin{aligned} = & 2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v +\int_t^0 \tan ^{(2 n-1)}(-z) d z \\ = & 2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v -\int_0^t \tan ^{(2 n-1)}(z) d z \\ = & 2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v-f(t) \\ = & f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v \\ = & f(t)+0\left[\because \int_0^\pi \tan ^{(2 n-1)}(v) d v=0\right] \\ & =f(t)+f(\pi) \quad[\because f(\pi)=0] \\ \therefore \quad & f(t+\pi)=f(t)+f(\pi) \end{aligned} $

$ \begin{aligned} I & =\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x \\ & =\int_0^2 x^8\left(\frac{4-x^2}{x^2}\right)^{5 / 2} d x \\ & =\int_0^2 x^8 \cdot \frac{\left(4-x^2\right)^{5 / 2}}{x^5} d x \\ & =\int_0^2 x^3\left(4-x^2\right)^{5 / 2} d x \end{aligned} $

Let $u=4-x^2$

Then, $d u=-2 x d x$, so $x d x=\frac{-1}{2} d u x^2=4-u$ and $x^3=x \cdot x^2=x(4-u)$ when $x=0, u=4$ and when $x=2, u=0$

$ \begin{aligned} & \therefore \int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x \\ & \quad=\int_0^2 x^3\left(4-x^2\right)^{5 / 2} d x \\ & \quad=\int_4^0(4-u) \cdot u^{5 / 2} \cdot\left(\frac{-1}{2}\right) d u \\ & \quad=\frac{1}{2}\left[4 \cdot \frac{u^{7 / 2}}{7 / 2}-\frac{u^{9 / 2}}{9 / 2}\right]_0^4 \\ & \quad=\frac{1}{2}\left[\frac{8}{7} u^{7 / 2}-\frac{2}{9} u^{9 / 2}\right]_0^4 \\ & \quad=\frac{1}{2}\left[\frac{8}{7}(u)^{7 / 2}-\frac{2}{9}(4)^{9 / 2}-0\right] \\ & \quad=\frac{1}{2}\left[\frac{8}{7}(128)-\frac{2}{9}(512)\right] \\ & \quad=\frac{1}{2}\left[\frac{1024}{7}-\frac{1024}{9}\right] \\ & \quad=\frac{1024}{2} \cdot \frac{2}{63}=\frac{1024}{63} \\ & \therefore \int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=\frac{1024}{63}=\frac{2^{10}}{63} \end{aligned} $

$ \begin{aligned} & \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cdot \cos ^2 x(\sin x+\cos x) d x \\ & =\int_{-\pi 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x \\ & \quad+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos ^3 x d x \\ & =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin x \cdot\left(1-\cos ^2 x\right) \cos ^2 x d x \\ & \quad+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x\left(1-\sin ^2 x\right) \cdot \sin ^2 x d x \end{aligned} $

$ \begin{gathered} =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin x \cdot \cos ^2 x \cdot d x \\ -\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin x \cdot \cos ^4 x d x \\ +\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x \cdot \sin ^2 x \cdot d x-\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x \cdot \sin ^4 x d x \\ =\left[\frac{\cos ^3 x}{3}+\frac{\cos ^5 x}{5}+\frac{\sin ^3 x}{3}-\frac{\sin ^5 x}{5}\right]_{-\frac{\pi}{2}}^\pi \end{gathered} $

$ \begin{aligned} & =\left(0+0+\frac{1}{3}-\frac{1}{5}\right)-\left(0+0-\frac{1}{3}+\frac{1}{5}\right) \\ & =\frac{2}{3}-\frac{2}{5}=\frac{10-6}{15}=\frac{4}{15} \end{aligned} $

$ \int_{\frac{1}{5}}^{\frac{1}{2}} \frac{\sqrt{x-x^2}}{x^3} d x $

$ =\int_{\frac{1}{5}}^{\frac{1}{2}} \frac{x \sqrt{\frac{1}{x}-1}}{x^3} d x \int {_{{1 \over 5}}^{{1 \over 2}}} {1 \over {{x^2}}}\sqrt {{1 \over x} - 1.dx}$

Put $\frac{1}{x}-1=t \Rightarrow \frac{1}{x^2} d x=-d t$

$ \begin{aligned} & =-\int_4^1 t^{\frac{1}{2}} \cdot d t=\int_1^4 \sqrt{t} \cdot d t \\ & =\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_1^4=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_1^4 \\ & =\frac{2}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right]=\frac{2}{3}(8-1) \\ & =\frac{14}{3} \end{aligned} $

$ \begin{aligned} I & =\int_0^{400 \pi} \sqrt{1-\cos 2 x} d x \\ & =\int_0^{400 \pi} \sqrt{2 \sin ^2 x} d x \\ & =\sqrt{2} \int_0^{400 \pi} \sin x d x \\ & =400 \sqrt{2} \int_0^\pi \sin x d x \\ & =2 \times 400 \sqrt{2} \int_0^{\pi / 2} \sin x d x \\ & =800 \sqrt{2}[-\cos x]_0^{\frac{\pi}{2}} \\ & =800 \sqrt{2}\left[-\cos \frac{\pi}{2}+\cos 0\right] \\ & =800 \sqrt{2} \end{aligned} $

Let $I=\int_0^x \frac{t^2}{\sqrt{a^2+t^2}} d t$

Put $t=a \sin h \theta \Rightarrow d t=a \cos h \theta d \theta$

Now, $\sqrt{a^2+t^2}=\sqrt{a^2+a^2 \sin h^2 \theta}$

$ =a \sqrt{1+\sinh ^2 \theta}=a \cos h \theta $

and   $ t^2=a^2 \sin h^2 \theta $

$ \begin{aligned} &\begin{array}{lll} \hline t & 0 & x \\ \hline \theta & 0 & \sin h^{-1}\left(\frac{x}{4}\right) \\ \hline \end{array}\\ &\begin{aligned} & \therefore I=\int_0^{\sin h^{-1}\left(\frac{x}{a}\right)} \frac{a^2 \sin h^2 \theta}{a \cos h \theta} \cdot a \cos h \theta d \theta \\ & \quad=\int_0^{\sin h^{-1}\left(\frac{x}{a}\right)} a^2 \sin h^2 \theta d \theta \\ & \because \sin h^2 \theta=\frac{\cos h(2 \theta)-1}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { So, } I=a^2 \int_0^{\sin h^{-1}\left(\frac{x}{a}\right)} \frac{\cos h(2 \theta)-1}{2} d \theta=\frac{a^2}{2} \\ & {\left[\begin{array}{c} \sin h^{-1}\left(\frac{x}{a}\right) \\ \int_0^{\cos h(2 \theta) d \theta-\int_0^{\sin h^{-1}\left(\frac{x}{a}\right)}} \int_1 1 \cdot d \theta \end{array}\right]} \\ & =\frac{a^2}{2}\left[\frac{1}{2} \sin h(2 \theta)-\theta\right]=\frac{a^2}{2} \\ & =\frac{a^2}{2}\left[\frac{x}{a} \times \frac{\sqrt{a^2+x^2}}{a}-\sin h^{-1}\left(\frac{x}{a}\right)\right] \\ & =\frac{x}{2} \sqrt{a^2+x^2}-\frac{a^2}{2} \sin h^{-1}\left(\frac{x}{a}\right) \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos ^4 x d x \\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\cos ^4 x} d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^4 x d x \\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^2 x\left(1+\tan ^2 x\right) d x=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\left(1+u^2\right) d u \end{aligned} $

$ \begin{array}{lll} \hline \boldsymbol{x} & \frac{\pi}{6} & \frac{\pi}{3} \\ \hline u & \frac{1}{\sqrt{3}} & \sqrt{3} \\ \hline \end{array} $

$ \begin{aligned} &\text { where } u=\tan x\\ &\begin{aligned} & =\left(u+\frac{u^3}{3}\right)_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \\ & =\left(\sqrt{3}+\frac{(\sqrt{3})^3}{3}\right)-\left(\frac{1}{\sqrt{3}}+\frac{\left(\frac{1}{\sqrt{3}}\right)^3}{3}\right) \\ & =\frac{3 \sqrt{3}+3 \sqrt{3}}{3}-\left(\frac{1}{\sqrt{3}}+\frac{1}{9 \sqrt{3}}\right) \\ & =\frac{6 \sqrt{3}}{3}-\left(\frac{9+1}{9 \sqrt{3}}\right)=2 \sqrt{3}-\frac{10 \sqrt{3}}{27} \\ & =\frac{54 \sqrt{3}-10 \sqrt{3}}{27}=\frac{44 \sqrt{3}}{27}=\frac{44}{9 \sqrt{3}} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} Let\,\,& I=\int_0^{\frac{3 \pi}{2}} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \quad \begin{aligned} I= & \int_0^{\frac{3 \pi}{2}} \frac{\left(\cos \left(\frac{3 \pi}{2}-x\right)\right)^3}{\left(\cos \left(\frac{3 \pi}{2}-x\right)\right)^3+\left(\sin \left(\frac{3 \pi}{2}-x\right)\right)^3} \end{aligned} d x \\ &=\int_0^{\frac{3 \pi}{2}} \frac{-\sin ^3 x}{-\sin ^3 x-\cos ^3 x} d x \\ &=\int_0^{\frac{3 \pi}{2}} \frac{\sin ^3 x}{\sin ^3 x+\cos ^3 x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \begin{aligned} &\text { On adding in Eqs. (i) and (ii), we get }\\ &\begin{array}{ll} \Rightarrow & 2 I=\int_0^{\frac{3 \pi}{2}} d x=\frac{3 \pi}{2} \\ \therefore & I=\frac{3 \pi}{4} \end{array} \end{aligned} $

$ \begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots+\frac{1}{k n}\right] \\ & \sum_{i=1}^{(k-1) n} \frac{1}{n+i} \\ & \Rightarrow \frac{1}{n} \sum_{i=1}^{(k-1) n} \frac{1}{1+\frac{i}{n}}=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{(k-1) n} \frac{1}{1+\frac{i}{n}} \\ & =\int_0^{k-1} \frac{1}{1+x} d x=[\ln (1+x)]_0^{k-1} \\ & =\ln k \end{aligned} $

$ \begin{aligned} & \text { } \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} d x \\ & \text { Let } x=\frac{5}{2} \sin \theta+\frac{3}{2} \\ & \Rightarrow d x=\frac{5}{2} \cos \theta d \theta \\ & \Rightarrow 4-x=4-\left(\frac{5}{2} \sin \theta+\frac{3}{2}\right) \\ & =\frac{5}{2}(1-\sin \theta) \\ & \Rightarrow x+1=\frac{5}{2} \sin \theta+\frac{3}{2}+1=\frac{5}{2}(1+\sin \theta) \\ & \Rightarrow \quad \frac{4-x}{x+1}=\frac{1-\sin \theta}{1+\sin \theta} \\ & \Rightarrow \quad \sqrt{\frac{4-x}{x+1}}=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{1-\sin \theta}{\cos \theta} \end{aligned} $

$ \begin{aligned} & \int \sqrt{\frac{4-x}{x+1}} d x=\int \frac{1-\sin \theta}{\cos \theta} \frac{5}{2} \cos \theta d \theta \\ & \quad=\frac{5}{2} \int(1-\sin \theta) d \theta \\ & \quad=\frac{5}{2}(\theta+\cos \theta)+c \\ & x=\frac{5}{2} \sin \theta+\frac{3}{2} \Rightarrow \sin \theta=\frac{2 x-3}{5} \end{aligned} $

$ \begin{gathered} \Rightarrow \cos \theta=\sqrt{1-\sin ^2 \theta}=\frac{2}{5} \sqrt{4+3 x-x^2} \\ \begin{aligned} \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} d x= & {\left[\frac{5}{2} \sin ^{-1}\left(\frac{2 x-3}{5}\right)\right.}+ \left.\sqrt{4+3 x-x^2}\right]_{-1}^4 \end{aligned} \\ =\frac{5}{2} \cdot \frac{\pi}{2}-\frac{5}{2}\left(\frac{-\pi}{2}\right) \\ =\frac{5 \pi}{4}+\frac{5 \pi}{4}=\frac{5 \pi}{2} \end{gathered} $

$ \text { } \begin{aligned} & \int_0^{\frac{\pi}{4}} \frac{\cos ^2 x d x}{\cos ^2 x+4 \sin ^2 x} \\ & \quad=\int_0^{\frac{\pi}{4}} \frac{1}{1+4 \tan ^2 x} d x \end{aligned} $

Put $u=2 \tan x$

$ \begin{aligned} d u & =2 \sec ^2 x d x \\ d x & =\frac{d u}{2\left(1+\tan ^2 x\right)}=\frac{2 d u}{4+u^2} \\ x & =0, u=0 \\ x & =\frac{\pi}{4}, u=2 \\ & =\int_0^2 \frac{1}{1+u^2} \cdot \frac{2 d u}{4+u^2} \\ & =\int_0^2\left(\frac{2}{3} \frac{1}{1+u^2}-\frac{2}{3} \frac{1}{4+u^2}\right) d u \\ & =\frac{2}{3}\left[\tan ^{-1} u\right]_0^2-\frac{2}{3} \cdot \frac{1}{2}\left[\tan ^{-1} \frac{u}{2}\right]_0^2 \\ & =\frac{2}{3} \tan ^{-1}(2)-\frac{\pi}{12} \end{aligned} $

$ \begin{aligned} & \text { } \int_{5 \pi}^{25 \pi} \sin 2 x+\cos 2 x \mid d x \\ & \Rightarrow|\sin 2 x+\cos 2 x|=\sqrt{2}\left|\sin \left(2 x+\frac{\pi}{4}\right)\right| \\ & \Rightarrow \int_{5 \pi}^{25 \pi} \sin 2 x+\cos 2 x \mid d x \\ & =\int_{5 \pi}^{25 \pi} \sqrt{2}\left|\sin \left(2 x+\frac{\pi}{4}\right)\right| d x \end{aligned} $

Let $u=2 x+\frac{\pi}{4}$

$ \begin{aligned} & d u=2 d x \\ & d x=\frac{1}{2} d u \\ & x=5 \pi, u=10 \pi+\frac{\pi}{4} \\ & x=25 \pi, u=50 \pi+\frac{\pi}{4} \\ & =\sqrt{2} \int_{10 \pi+\frac{\pi}{4}}^{50 \pi+\frac{\pi}{4}}|\sin u| \frac{1}{2} d u \\ & =\frac{\sqrt{2}}{2} \int_{10 \pi+\frac{\pi}{4}}^{50 \pi+\frac{\pi}{4}}|\sin u| d u \end{aligned} $

The period of $|\sin u|$ is $\pi$.

The length of the interval of integration is

$ \begin{aligned} & =\left(50 \pi+\frac{\pi}{4}\right)-\left(10 \pi+\frac{\pi}{4}\right)=40 \pi \\ & =\frac{\sqrt{2}}{2} \cdot 40 \int_0^\pi|\sin u| d u \\ & =\frac{\sqrt{2}}{2} \times 40 \times 2 \\ & =40 \sqrt{2} \end{aligned} $

$I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{3}}\left|\tan \left(x-\frac{\pi}{6}\right)\right| d x$

Put $\tan \left(x-\frac{\pi}{6}\right)=0 \Rightarrow x-\frac{\pi}{6}=0$

$ \begin{aligned} & I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{6}}-\tan \left(x-\frac{\pi}{6}\right) d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \tan \left(x-\frac{\pi}{6}\right) d x \\ & \begin{array}{r} =-\left[\log \left(\sec \left(x-\frac{\pi}{6}\right)\right]_{-\pi}^{\frac{\pi}{6}}\right. +\left[\log \sec \left(x-\frac{\pi}{6}\right)\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \end{array} \\ & \Rightarrow \log \left(\sec \left(\frac{5 \pi}{12}\right)\right)+\log \left(\sec \left(\frac{\pi}{6}\right)\right) \\ & \quad=\log \left(\sec \left(\frac{5 \pi}{12}\right) \sec \left(\frac{\pi}{6}\right)\right) \\ & =\log \left(2 \sqrt{2} \frac{(\sqrt{3}+1)}{\sqrt{3}}\right) \end{aligned} $

$ \begin{aligned} \text { } I= & \int_0^\pi \frac{x \sin x}{\sin ^2 x+2 \cos ^2 x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow I= & \int_0^\pi \frac{(\pi-x) \sin x}{\sin ^2 x+2 \cos ^2 x} d x \\ & \quad\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow 2 I=\int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x \\ & \text { Put } \cos x=t \Rightarrow-\sin x d x=d t \\ & \Rightarrow 2 I=-\pi\left[\tan ^{-1} t\right]_1^{-1} \\ & \Rightarrow \quad 2 I=-\pi\left[\frac{-\pi}{4}-\frac{\pi}{4}\right] \Rightarrow I=\frac{\pi^2}{4} \end{aligned} $

$ \text { } \mathop {\lim }\limits_{n \to \infty }\left[\frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\frac{n}{n^2+n^2}\right] $

$ \begin{aligned} & \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r}{r^2+n^2}=\lim _{n \rightarrow \infty} \sum \frac{\left(\frac{r}{n}\right)}{1+\left(\frac{r}{n}\right)^2} \cdot \frac{1}{n} \\ & \Rightarrow \int_0^1 \frac{x}{1+x^2} d x \\ & \text { put } 1+x^2=t \Rightarrow 2 x d x=d t \\ & \Rightarrow \frac{1}{2} \int_1^2 \frac{1}{t} d t \Rightarrow \frac{1}{2}[\log t]_1^2=\frac{1}{2} \log 2 \end{aligned} $

$ \begin{aligned} & \int_0^{\frac{\pi}{2}} \ln (\tan x+\cot x) d x=\int_0^{\frac{\pi}{2}} \ln \left(\frac{1}{\sin x \cos x}\right) d x \\ & =\int_0^{\frac{\pi}{2}}(\ln 2-\ln \sin 2 x) d x \\ & =\ln 2 \int_0^{\frac{\pi}{2}} d x-\int_0^{\frac{\pi}{2}} \ln \sin 2 x d x \\ & \text { put } 2 x=t, d x=\frac{d t}{2} \\ & =\frac{\pi}{2} \ln 2-\frac{1}{2} \int_0^\pi \ln \sin t d t \\ & =\frac{\pi}{2} \ln 2-\int_0^{\pi / 2} \ln \sin t d t \\ & =\frac{\pi}{2} \ln 2-\left(-\frac{\pi}{2} \ln 2\right) \end{aligned} $

$ \left[\because \int_0^{\pi / 2} \ln \sin x=-\frac{\pi}{2} \ln 2\right] $

$ =\pi \ln 2 $

Let $I=\int_0^\pi x \sin ^5 x \cos ^6 x d x$

$ \begin{aligned} &\begin{aligned} \Rightarrow \quad I & =\frac{\pi}{2} \int_0^\pi \sin ^5 x \cos ^6 x d x \\ & =\frac{\pi}{2} 2 \int_0^{\pi / 2} \sin ^5 x \cos ^6 x d x \end{aligned}\\ &\text { By Walli's Formula, }\\ &=\pi\left[\frac{(4 \cdot 2)(5 \cdot 3 \cdot 1)}{11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1}\right]=\frac{8 \pi}{693} \end{aligned} $

$ \begin{aligned} & \text { } I=\int_{1 / 2}^{1 / \sqrt{2}} \frac{1}{\left(x+\sqrt{1-x^2}\right)\left(1-x^2\right)} d x \\ & I=\int_{\pi / 6}^{\pi / 4} \frac{\cos \theta d \theta}{(\sin \theta+\cos \theta) \cos ^2 \theta} \\ & \text { Put } x=\sin \theta \Rightarrow d x=\cos \theta d \theta \\ & \quad=\int_{\pi / 6}^{\pi / 4} \frac{d \theta}{\sin \theta \cos \theta+\cos ^2 \theta} \end{aligned} $

Dividing $N^T$ and $D^r$ by $\cos ^2 \theta$, we get

$ =\int_{\pi / 6}^{\pi / 4} \frac{\sec ^2 \theta}{\tan \theta+1} d \theta $

Put $\tan \theta+1=t, \sec ^2 \theta d \theta=d t$

$ \begin{aligned} & =\int_{\frac{1}{\sqrt{3}}+1}^2 \frac{d t}{t}=[\ln t]_{1+\frac{1}{\sqrt{3}}}^2 \\ & =\ln 2-\ln \left(1+\frac{1}{\sqrt{3}}\right) \\ & =\ln 2-\ln \left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)=\ln \left(\frac{2 \sqrt{3}}{\sqrt{3}+1}\right) \\ & =\ln \left(\frac{2 \sqrt{3}(\sqrt{3}-1)}{3-1}\right)=\ln (3-\sqrt{3}) \end{aligned} $

$ \begin{aligned} &\text { Given, } H(x)=3 x^4+6 x^3-2 x^2+1\\ &\begin{aligned} & \text { Given, } \frac{H(x)}{(x-1)(x-2)(x+1)} \\ & \begin{aligned} &=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2} \text { taking LCM } \\ & H(x)=f(x)(x-1)(x+1)(x-2)+g(x) \\ & H(-1)=g(-1) \\ & H(2)=g(2) \\ & H(1)=g(1) \\ & \text { Now, } H(-1)+2 H(2)-3 H(1) \\ &=g(-1)+2 g(2-3 g(1) \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \begin{aligned} I & =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\cos x-\sin x}{\sin 2 x} d x \\ I & =\int_{\sqrt{2}}^{\frac{\sqrt{3}}{2}+\frac{1}{2}} \frac{d t}{t^2-1} \end{aligned}\\ &\text { Put } \sin x+\cos x=t\\ &\begin{aligned} & (\cos x-\sin x) d x=d t \\ & 1+\sin 2 x=t^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & =\left[\frac{1}{2} \ln \left(\frac{t-1}{t+1}\right)\right]_{\sqrt{2}}^{\frac{\sqrt{3}+1}{2}} \\ & =\frac{1}{2}\left[\ln \left(\frac{\frac{\sqrt{3}+1}{2}-1}{\frac{\sqrt{3}+1}{2}+1}\right)-\ln \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right] \\ & =\frac{1}{2}\left[\ln \frac{\sqrt{3}-1}{\sqrt{3}+3}-\ln \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right] \\ & =\frac{1}{2}\left[\ln \frac{2(2 \sqrt{3}-3)}{6}-\ln (3-2 \sqrt{2})\right] \\ & =\frac{1}{2}\left[\ln \frac{2 \sqrt{3}-3}{3}+\ln (3+2 \sqrt{2})\right] \\ & =\frac{1}{2}\left[\ln \left(\frac{2-\sqrt{3}}{\sqrt{3}}\right)+\ln (3+2 \sqrt{2})\right] \\ & =\frac{1}{2} \ln \left[\frac{(3+2 \sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right] \end{aligned} $

$ \text { } \begin{aligned} I & =\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos x+\sin x} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x \end{aligned} $

$ \begin{aligned} & =\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\sin \frac{x}{2}+\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x \\ & =\frac{1}{2} \int_0^{\frac{\pi}{2}}\left(1+\frac{\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) d x \\ & =\frac{1}{2}\left[x+\left\{-2 \ln \left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\right\}\right]_0^{\frac{\pi}{2}} \\ & =\frac{1}{2}\left[x-2 \ln \left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\right]_0^{\frac{\pi}{2}} \\ & =\frac{1}{2}\left[\left(\frac{\pi}{2}-2 \ln \sqrt{2}\right)-0\right] \\ & =\frac{\pi}{4}-\ln \sqrt{2}=\frac{\pi}{4}-\frac{1}{2} \ln 2 \end{aligned} $

$I=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Apply property

$ \begin{aligned} & \int_a^b x f(x) d x=\frac{a+b}{2} \int f(x) d x \\ & \text { If } f(a+b-x)=f(x) \\ & \qquad I=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x \end{aligned} $

$ \begin{aligned} & \text { Put } \cos x=t \\ & \sin x d x=-d t \\ & \qquad \begin{aligned} I & =-\frac{\pi}{2} \int_{+1}^{-1} \frac{d t}{1+t^2} \\ & =\frac{\pi}{2} \int_{-1}^1 \frac{d t}{1+t^2}=\frac{\pi}{2} \cdot 2 \int_0^1 \frac{d t}{1+t^2} \\ & =\pi\left(\tan ^{-1} t\right)_0^1 \\ & =\pi \tan ^{-1} 1=\pi \times \frac{\pi}{4}=\frac{\pi^2}{4} \end{aligned} \end{aligned} $

Let $I=\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x$

As $\sin ^4 x+\cos ^4 x$ is periodic with $\frac{\pi}{2}$

So, $\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x$

$ \begin{aligned} & =\int_0^{4 \times \frac{\pi}{2}}\left(\sin ^4 x+\cos ^4 x\right) d x \\ = & 4 \int_0^{\pi / 2}\left(\sin ^4 a x+\cos ^4 x\right) d x \\ = & 4 \int_0^{\pi / 2} 1-\frac{\sin ^2 2 x}{2} d x \\ = & 4 \int_0^{\pi / 2} \frac{4-1+\cos 4 x}{4} d x \\ = & 4 \int_0^{\pi / 2} \frac{3+\cos 4 x}{4} d x \\ = & {\left[3-\frac{\sin 4 x}{4}\right]_0^{\frac{\pi}{2}} } \\ = & 3\left[\frac{\pi}{2}\right]=\frac{3 \pi}{2} \end{aligned} $

Now, $\int_0^\pi \sin ^2 x d x=2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x$

$ \begin{aligned} & =2 \int_0^{\frac{\pi}{2}}\left(\frac{1-\cos 2 x}{2}\right) d x=\left[x-\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}} \\ & =\frac{\pi}{2} \end{aligned} $

Similarly,

$ \begin{aligned} & \int_0^{\frac{\pi}{2}} \cos ^2 x d x=\frac{\pi}{4} \\ \therefore & \frac{3 \pi}{2}=K \times \frac{\pi}{2}+L \times \frac{\pi}{4} \\ & 2 K+L=6 \end{aligned} $

If $K=1$, then $L=4$ and if $K=2$, then $L=2$

$\therefore$ There are two possible ordered pair i.e. $(1,4),(2,2)$.

$ \text { } \begin{aligned}Let \,\, I & =\int_0^\pi \frac{x \sin x d x}{4 \cos ^2 x+3 \sin ^2 x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & =\int_0^\pi \frac{x \sin x d x}{4 \cos ^2 x+3-3 \cos ^2 x} \\ & =\int_0^\pi \frac{x \sin x}{\cos ^2 x+3} \\ & =\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{\cos ^2(\pi-x)+3} \end{aligned} $

$ \begin{aligned} &=\int_0^\pi \frac{(\pi-x) \sin x}{\cos ^2 x+3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { On adding Eq. (i) and (ii), we get }\\ &\begin{aligned} 2 I & =\int_0^\pi \frac{\pi \sin x}{3+\cos ^2 x} d x \\ & =\pi \int_0^\pi \frac{\sin x}{3+\cos ^2 x} d x=-\pi \int_1^{-1} \frac{1}{3+t^2} d t \\ & =-\pi \frac{1}{\sqrt{3}}\left(\tan ^{-1} \frac{t}{\sqrt{3}}\right)_1^{-1} \\ \Rightarrow \quad 2 I & =\frac{-\pi}{\sqrt{3}}\left[\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right] \\ & =\frac{-\pi}{\sqrt{3}}\left[-\frac{\pi}{6}-\frac{\pi}{6}\right] \\ & =-\frac{\pi}{\sqrt{3}} \times\left(-\frac{2 \pi}{6}\right)=\frac{\pi^2}{3 \sqrt{3}} \\ \Rightarrow \quad I & =\frac{\pi^2}{6 \sqrt{3}} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int\left(\frac{1+x^2}{1+x^4}\right) d x \\ & =\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x^2}-2+2\right)} d x \\ & =\int \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2} d x \end{aligned} $

Let $x-\frac{1}{x}=t$

$ \begin{aligned} & \Rightarrow\left(1+\frac{1}{x^2}\right) d x=d t \\ & =\int \frac{1}{t^2+2} d t=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right) \\ & =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right) \text { or } \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right) \end{aligned} $

$ \begin{aligned} A & =\int_0^{\infty} \frac{1+x^2}{1+x^4} d x=\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)\right]_0^{\infty} \\ & =\frac{1}{\sqrt{2}}\left[\tan ^{-1} \infty-\tan ^{-1}(-\infty)\right] \\ & =\frac{1}{\sqrt{2}} \times 2 \times \frac{\pi}{2} \Rightarrow \frac{\pi}{\sqrt{2}} \\ B & =\int_0^1 \frac{1+x^2}{1+x^4} d x=\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)\right]_0^1 \\ & =\frac{1}{\sqrt{2}}\left[\tan ^{-1}(0)-\tan ^{-1}(-\infty)\right]=\frac{1}{\sqrt{2}} \times \frac{\pi}{2} \\ & \Rightarrow A=2 B \end{aligned} $

We have, $I=\int_0^1 \sqrt{\frac{2+x}{2-x}} d x$

Let $\frac{2+x}{2-x}=u$

$ \left(\frac{1}{2-x}+\frac{x+2}{(2-x)^2}\right) d x=d u $

Now, lower limit $=1$

and upper limit $=3$

$ =\int_1^3 \frac{\sqrt{u}}{(u+1)^2} d u $

Now, let $s=\sqrt{u}$

$ \begin{aligned} d s & =\frac{1}{2 \sqrt{u}} \cdot d u=8 \int_1^{\sqrt{3}} \frac{s^2}{\left(s^2+1\right)^2} d s \\ & =8 \int_1^{\sqrt{3}}\left(\frac{1}{s^2+1}-\frac{1}{\left(s^2+1\right)^2}\right) d s \\ & =8\left|\tan ^{-1}(s)\right|_1^{\sqrt{3}}-8 \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s \\ & =8\left(\tan ^{-1}(\sqrt{3})-\tan ^{-1}(1)\right)-8 \end{aligned} $

$ \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s $

$ \begin{aligned} & =8\left(\frac{\pi}{3}-\frac{\pi}{4}\right)-8 \int_1^{\sqrt{3}} \frac{1}{\left(1+s^2\right)^2} d s \\ & =\frac{2 \pi}{3}-8 \int_{\pi / 4}^{\pi / 3} \cos ^2 p d p \end{aligned} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { [put } s=\tan p] $

$ \begin{aligned} & =\frac{2 \pi}{3}+2-\sqrt{3}+1-\left.4 p\right|_{\pi / 4} ^{\pi / 3} \\ & =\frac{1}{3}(6-3 \sqrt{3}+\pi)=\frac{\pi}{3}+2-\sqrt{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & M=\int_0^{\infty} \frac{\log t}{1+t^3} \text { and } N=\int_{-\infty}^{\infty} \frac{t e^{2 t}}{1+e^{3 t}} d t \\ & N=\int_{-\infty}^{\infty} \frac{t \cdot e^{2 t}}{1+e^{3 t}} d t \end{aligned} \end{aligned} $

Let $e^t=x$

$e^t \cdot d t=d x$

Lower limit = 0

and Upper limit $=\infty$

$ \begin{aligned} N & =\int_0^{\infty} \frac{\log x \cdot x^2}{1+x^3} \cdot \frac{d x}{x}=\int_0^{\infty} \frac{x \log x}{1+x^3} d x \\ & =\int_0^{\infty} \frac{x \log x}{1+x^3} d x=-\int_0^{\infty} \frac{\log x}{1+x^3} d x \\ \Rightarrow N & =-\int_0^{\infty} \frac{\log t}{1+t^3} d t \Rightarrow N=-M \end{aligned} $

To solve the integral $ I = \int_{-2}^2 \left(4 - x^2\right)^{\frac{5}{2}} \, dx $, we can use a substitution method.

Substitution:

Substitute $ x = 2 \sin u $:

Then $ dx = 2 \cos u \, du $.

Change the limits:

When $ x = -2 $, $ u = -\frac{\pi}{2} $.

When $ x = 2 $, $ u = \frac{\pi}{2} $.

Substitute into the integral:

$ I = \int_{-\pi/2}^{\pi/2} \left(4 - (2\sin u)^2\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

$ = \int_{-\pi/2}^{\pi/2} \left(4 - 4\sin^2 u\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

Simplify:

Since $ 4 - 4\sin^2 u = 4\cos^2 u $, we have:

$ = \int_{-\pi/2}^{\pi/2} \left(4\cos^2 u\right)^{\frac{5}{2}} \cdot 2\cos u \, du $

$ = 32 \int_{-\pi/2}^{\pi/2} \cos^6 u \, du $

Using Reduction Formula:

Solving this integral using the reduction formula yields:

$ I = 64 \int_{-\pi/2}^{\pi/2} \cos^6 u \, du = 20\pi $

Thus, the value of the integral is $ I = 20\pi $.

$ \begin{aligned} &\text { We have, } P=\lim _{x \rightarrow \infty}\\ &\begin{aligned} & \quad\left[\left(1+\frac{1}{n^3}\right)^{1 / n^3}-\left(1+\frac{8}{n^3}\right)^{4 / n^3} \ldots(2)^{1 / n}\right] \\ & \Rightarrow \quad \log p=\lim _{x \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{r^2}{n^2} \log \left(1+\frac{r^3}{n^3}\right) \\ & \Rightarrow \quad \log p=\int_0^1 x^2 \log \left(1+x^3\right) d x \\ & \text { Put } 1+x^3=t \Rightarrow 3 x^2 d x=d t \\ & \text { when } n=0, t=1 \text { and } x=1, t=2 \\ & \therefore \log p=\frac{1}{3} \int_1^2 \log t d t=\frac{1}{3}[t \log t-t]_1^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \log p=\frac{1}{3}[2 \log 2-1] \\ & \text { Hence, } p=e^{(2 \log 2-1 / 3)} \end{aligned} $

To solve the integral $ \int_{-5 \pi}^{5 \pi}(1-\cos 2 x)^{\frac{5}{2}} d x $, we start by simplifying the expression inside the integral:

Notice that $ 1 - \cos 2x = 2 \sin^2 x $. Substituting this in, the integral becomes:

$ I = \int_{-5 \pi}^{5 \pi} (1 - \cos 2x)^{\frac{5}{2}} \, dx = \int_{-5 \pi}^{5 \pi} (2 \sin^2 x)^{\frac{5}{2}} \, dx $

Simplify the expression:

$ = \int_{-5 \pi}^{5 \pi} 2^{\frac{5}{2}} \cdot |\sin^5 x| \, dx $

Since $|\sin^5 x| = \sin^5 x$ for the interval $[0, \pi]$ and is symmetrical over $[-\pi, 0]$, the integral becomes:

$ = 8 \sqrt{2} \int_0^{5\pi} \sin^5 x \, dx $

Exploit the periodicity of the sine function to break the integral over one full period:

$ = 40 \sqrt{2} \int_0^{\pi} \sin^5 x \, dx $

Using the reduction formula for $\int_0^\pi \sin^n x \, dx$, the final result of this reduction gives:

$ = \frac{128 \sqrt{2}}{3} $

$I=\int_0^{\pi / 4} \log (1+\tan x) d x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using $\int_0^a f(x) d x=\int_0^a f(a-x) d x$

$ \begin{aligned} I & =\int_0^{\pi / 4} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\ & =\int_0^{\pi / 4} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x \\ & \left(\because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\right) \\ I & =\int_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On adding Eq. (i) and Eq. (ii), we get

$ \begin{aligned} 2 I= & \int_0^{\pi / 4} \log (1+\tan x) d x \\ & \quad+\int_0^{\pi / 4} \log \frac{2}{(1+\tan x)} d x \\ = & \int_0^{\pi / 4} \log (1+\tan x) \cdot \frac{2}{(1+\tan x)} d x \\ = & \int_0^{\pi / 4} \log 2 d x=\log 2 \int_0^{\pi / 4} d x \\ = & \log 2 \cdot(x)_0^{\pi / 4}=\log 2 \cdot\left(\frac{\pi}{4}-0\right) \\ 2 I= & \frac{\pi}{4} \log 2 \Rightarrow I=\frac{\pi}{8} \log 2 \end{aligned} $

$ \begin{aligned} I & =\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x \\ & =2 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =2\left[\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x\right] \\ & =2\left[\pi \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x}-\int_0^\pi \frac{x \sin x}{1+\cos ^2 x}\right] \\ 2 I & =2 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x \end{aligned}\\ &\text { }\\ &\begin{aligned}Let\,\,\,\,\, t & =\cos x \\ d t & =-\sin x d x \\ & =-\pi \int_1^{-1} \frac{1}{1+t^2} d t=\pi \int_{-1}^1 \frac{1}{1+t^2} d t \\ I & =\pi\left[\tan ^{-1}(t)\right]_{-1}^1 \\ & =\pi\left[\frac{\pi}{4}+\frac{\pi}{4}\right]=\frac{\pi^2}{2} \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} & \quad I=\int_0^{\pi / 4} \frac{x^2 d x}{(x \sin x+\cos x)} \Rightarrow I=\int_0^{\pi / 4} \frac{x \sec x \cdot \cos x \cdot d x}{(x \sin x+\cos x)^2} \\ & \Rightarrow \quad I=\int_0^{\pi / 4} x \sec x \cdot \frac{x \cos x}{(x \sin x+\cos x)^2} d x \end{aligned} $

Let $I_1=x \sec x$ and $I_2=\frac{x \cos x}{x \sin x+\cos x}$

Let $x \sin x+\cos x=t$

$ \begin{aligned} & x \cos x+\sin x-\sin x=\frac{d t}{d x} \\ & x \cos x d x=d t \end{aligned} $

$ \begin{array}{ll} \Rightarrow & I_2=\int \frac{x \cos x}{(x \sin x+\cos x)^2}=\int \frac{d t}{t^2}=-\frac{1}{t} \\ \Rightarrow \quad & I_2=\frac{-1}{(x \sin x+\cos x)} \\ \Rightarrow \quad & I=\int_0^{\pi / 4} x \sec x \cdot \frac{x \cos x}{(x \sin x+\cos x)} d x \\ \Rightarrow \quad & I=\left[x \sec x \cdot \frac{-1}{(x \sin x+\cos x)}\right]_0^{\frac{\pi}{4}}- \int_0^{\pi / 4}[\sec x+x \sec x \tan x] \cdot \frac{-1}{(x \sin x+\cos x)} d x \\ \because \quad & \int f(x) g(x) d x=f(x) \int g(x) d x-\int\left[\frac{d}{d x} f(x)-\int g(x) d x\right] d x \end{array} $

$ \begin{array}{ll} \Rightarrow & I=\left[\frac{-x \sec x}{x \sin x+\cos x}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \frac{\sec x(1+x \tan x) d x}{(x \sin x+\cos x)} \\ \Rightarrow & I=\left[\frac{-x \sec x}{x \sin x+\cos x}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \frac{\sec ^2 x(\cos x+x \sin x)}{(x \sin x+\cos x)} d x \\ \Rightarrow & I=\left[\frac{-x \sec x}{(x \sin x+\cos x)}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \sec ^2 x d x \end{array} $

$ \begin{array}{ll} \Rightarrow & I=\frac{-\frac{\pi}{4} \cdot \sqrt{2}}{\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}+\int_0^{\pi / 4} \sec ^2 x d x \\ \Rightarrow & I=\frac{-\pi \sqrt{2} / 4}{(\pi+4) / 4 \sqrt{2}}+[\tan x]_0^{\pi / 4} \\ \Rightarrow & I=\frac{-2 \pi}{\pi+4}+(1-0)=\frac{-2 \pi}{\pi+4}+1 \Rightarrow I=\frac{4-\pi}{\pi+4} \end{array} $

We have, $I=\int_0^1 \frac{x}{(1-x)^{\frac{3}{4}}} d x$

Let $1-x=t \Rightarrow x=1-t \Rightarrow-d x=d t$

$ \begin{aligned} & \therefore \quad I=\int_0^1 \frac{1-t}{t^{3 / 4}} d t \Rightarrow I=\int_0^1 t^{-3 / 4}(1-t) d t \\ & \Rightarrow \quad I=\int_0^1 t^{-3 / 4} d t-\int_0^1 t^{1 / 4} d t \Rightarrow I=\left[4 t^{1 / 4}\right]_0^1-\left[\frac{4 t^{5 / 4}}{5}\right]_0^1 \end{aligned} $

Putting the value of $t$

$ I=[4(1-x)]_0^1+\left[\frac{4}{5}(1-x)^{5 / 4}\right]_0^1 $

On solving, we get

$ I=\frac{16}{5} $

We have,

$ I=\int_{-1}^1\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) d x $

Let $f(x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$

$ \begin{aligned} & \Rightarrow f(-x)=\sqrt{1-x+x^2}-\sqrt{1+x+x^2} \\ & \Rightarrow f(-x)=-\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) \Rightarrow f(-x)=-f(x) \end{aligned} $

$\because f(x)$ is an odd function.

As, we know that

$ \int_{-a}^a f(x) d x=\left\{\begin{array}{l} \int_0^{2 a} f(x) d x, \text { if } f(-x)=f(x) \text {, even function } \\ 0, \text { if } f(-x)=-f(x), \text { odd funciton } \end{array}\right\} $

$ \begin{aligned} & \text { Then, } I=\int_{-1}^1 \sqrt{1+x+x^2} \\ & \qquad-\sqrt{1-x+x^2} d x=0 \end{aligned} $

We have,

$ \Rightarrow \quad I=\int_1^5|x-3|+|1-x| d x $

Let $f(x)=|x-3|+|1-x|$

Here, when $1 \leq x \leq 3$

$ \Rightarrow f(x)=-x-3-1+x=2 $

When, $3 \leq x \leq 5$

$ \begin{aligned} \Rightarrow f(x) & =x-3-1+x=2 x-4 \\ \Rightarrow \quad I & =\int_1^5|x-3|+|1-x| d x \\ & =\int_1^3 2 d x+\int_3^5(2 x-4) d x \\ \Rightarrow \quad I & =[2 x]_1^3+\left[\frac{2 x^2}{2}-4 x\right]_3^5 \end{aligned} $

On solving, we get

$ I=12 $

If $729 \int_1^3 \frac{1}{x^3\left(x^2+9\right)^2} d x=a+\log b$  Then, $a-b$

On substituting $u=x^2$

$ \begin{array}{ll} \Rightarrow & d u=2 x d x \\ \Rightarrow & d x=\frac{1}{2} \frac{d u}{\sqrt{u}} \\ \text { when, } x=1 & \end{array} $

$ \begin{array}{ll} \Rightarrow & u=1 \\ & x=3 \\ \Rightarrow & u=9 \end{array} $

$ \begin{aligned} &\begin{aligned} I & =729 \int_1^9 \frac{1}{u^{\frac{3}{2}}(u+9)^2} d u \\ & =729 \int_1^9 \frac{1}{2 u^{\frac{3}{2}+\frac{1}{2}}(u+9)^2} d u \\ & =729 \int_1^9 \frac{1}{2 u^2(u+9)^2} d u \end{aligned}\\ &\text { Partial fraction }\\ &\frac{1}{u^2(u+9)^2}=\frac{A}{u}+\frac{B}{u^2}+\frac{C}{(u+9)}+\frac{D}{(u+9)} \end{aligned} $

$ \begin{aligned} &\begin{gathered} \frac{1}{u^2(u+9)^2} \frac{A\left(u(u+9)^2+B(u+9)^2+C u^2(u+9)+D u^2\right.}{u^2(u+9)^2} \\ 1=A u\left(u^2+81+18 u\right)+B\left(u^2+81+18 u\right)+C u^3+9 C u^2+D u^2 \\ 1=A u^3+81 A u+18 A u^2+B u^2+81 B+18 u B+C u^3+9 C u^2+D u^2 \\ 1=(A+C) u^3+(18 A+B+9 C+D) u^2+(81 A+18 B) u+81 B \end{gathered}\\ &\text { Comparison of coefficient } u^3, u^2, u \text { and } 1 \text {. }\\ &\begin{aligned} & A+C=0,18 A+B+9 C+D=0 \\ & 81 A+18 B=0 \Rightarrow 81 B=1 \\ & B=\frac{1}{81} \Rightarrow 81 A+18 \cdot \frac{1}{81}=0 \\ & 81 A=-\frac{2}{9} \Rightarrow A=\frac{-2}{729} \\ & \frac{-2}{729}+C=0 \Rightarrow C=\frac{2}{729} \\ & 18\left(\frac{-2}{729}\right)+\frac{1}{81}+9\left(\frac{2}{729}\right)+D=0 \\ & -\frac{4}{81}+\frac{1}{81}+\frac{2}{81}+D=0 \Rightarrow \frac{-1}{81}+D=0 \end{aligned} \end{aligned} $

$ \begin{aligned} D & =\frac{1}{81} \\ & =\frac{729}{2} \int_1^9\left(\left(\frac{-2}{729}\right) \frac{1}{u}+\frac{1}{81 u^2}+\frac{2}{729(u+9)}+\frac{1}{81(u+9)^2}\right) d u \\ & =\frac{729}{2}\left[\frac{-2}{729} \log u+\frac{1}{81} \frac{u^{-2+1}}{(-2+1)}+\frac{2}{729} \log (u+9)+\frac{1}{81} \frac{(u+9)^{-2+1}}{-2+1}\right]_1^9 \\ & =\frac{729}{2}\left[\frac{-2}{729}(\log 9-\log 1)+\frac{1}{81}(-1)\left(\frac{1}{9}-1\right)+\frac{2}{729}[\log (18)-\log (10)]\right. \\ & =\frac{729}{2}\left[\frac{-2}{729} \log 9-\frac{1}{81}\left(\frac{-8}{9}\right)+\frac{2}{729} \log \left(\frac{18}{10}\right)-\frac{1}{81}\left(\frac{-8}{180}\right)\right]+C \\ & =\frac{1}{2}\left[-2 \log 9+8+2 \log \left(\frac{1}{5}\right)+\frac{8}{20}\right]+C \\ & =\frac{1}{2}\left[-2 \log 9+8+2 \log 9-2 \log 5+\frac{2}{5}\right]+C \\ & =\frac{1}{2}\left[8+\frac{2}{5}-2 \log 5\right]+C \\ & =\frac{1}{2}\left[\frac{40+2}{5}-2 \log 5\right] \Rightarrow \frac{42}{2 \times 5}-\log 5 \end{aligned} $

$ \begin{aligned} a & =\frac{42}{2 \times 5}, b=\frac{1}{5} \\ a & =\frac{21}{5}, b=\frac{1}{5} \\ a-b & =\frac{21}{5}-\frac{1}{5}=\frac{20}{5}=4 \end{aligned} $

To find the limit

$ \lim\limits_{n \rightarrow \infty} \frac{1^{77}+2^{77}+\ldots+n^{77}}{n^{78}} $

we use the method of integral approximation to approximate the sum of powers. Consider the sum

$ S_n = 1^{77} + 2^{77} + \ldots + n^{77}. $

This can be approximated using the integral

$ S_n \approx \int_1^n x^{77} \, dx. $

Evaluating this integral, we have

$ \int_1^n x^{77} \, dx = \left[\frac{x^{78}}{78}\right]_1^n = \frac{n^{78}}{78} - \frac{1}{78}. $

As $ n $ becomes very large, the term $\frac{1}{78}$ becomes negligible, and we approximate

$ S_n \approx \frac{n^{78}}{78}. $

Thus, the given limit can be rewritten and evaluated as follows:

$ \lim_{n \rightarrow \infty} \frac{S_n}{n^{78}} = \lim_{n \rightarrow \infty} \frac{\frac{n^{78}}{78}}{n^{78}} = \lim_{n \rightarrow \infty} \frac{1}{78} = \frac{1}{78}. $

Therefore, the value of the limit is

$ \frac{1}{78}. $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } f(x)=\left\{\begin{array}{cc} \frac{6 x^2+1}{4 x^3+2 x+3}, & 0< x < 1 \\ x^2+1 & 1 \leq x < 2 \end{array}\right. \\ & \int_0^2 f(x) d x=\int_0^1\left(\frac{6 x^2+1}{4 x^3+2 x+3}\right) d x+\int_1^2\left(x^2+1\right) d x \\ & \text { Let } u=4 x^3+2 x+3 \\ & \qquad d u=\left(12 x^2+2\right) d x \Rightarrow 2\left(6 x^2+1\right) d x \\ & \Rightarrow d x=\frac{d u}{2\left(6 x^2+1\right)} \\ & \quad x=0 \Rightarrow u=3 \\ & \quad x=1 \Rightarrow u=4+2+3=9 \\ & \text { Or }\left(6 x^2+1\right) d x=\frac{d u}{2} \Rightarrow \int_3^9 \frac{1}{u}\left(\frac{d u}{2}\right)+\int_1^2\left(x^2+1\right) d x \end{aligned} \end{aligned} $

$ \begin{aligned} & \frac{1}{2}[\log u]_3^9+\left[\frac{x^3}{3}+x\right]_1^2 \\ & \frac{1}{2}[\log 9-\log 3]+\left[\frac{8}{3}+2-\frac{1}{3}-1\right] \\ & \frac{1}{2}\left[\log \left(\frac{9}{3}\right)+\left(\frac{8+6}{3}-\frac{4}{3}\right)\right] \\ & \frac{1}{2} \log 3+\frac{14}{3}-\frac{4}{3}=\frac{1}{2} \log 3+\frac{10}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \int_1^n[x] d x=120 \\ & =\int_1^2 1^{d x}+\int_2^3 2^{d x}+\ldots+\int_{n-1}^n(n-1) d x=120 \\ & \Rightarrow \quad[x]_1^2+[2 x]_2^3+\ldots[(n-1) x]_{n-1}^n=120 \\ & \Rightarrow \quad 2-1+6-4+\ldots+(n-1)=120 \\ & \Rightarrow \quad 1+2+\ldots+n-1=120 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{n(n-1)}{2}=120 \\ & \Rightarrow \quad n^2-n-240=0 \\ & \Rightarrow \quad n^2-16 n+15 n-240=0 \\ & \Rightarrow \quad n(n-16)+15(n-16)=0 \\ & (n-16)(n+15)=0 \\ & n=16 \end{aligned} $