Definite Integration · Mathematics · AP EAPCET

$$ \int_0^1 x \sin ^{-1} x d x= $$

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

Here $[x]$ is the greatest integer function

$$ \int_0^2 x^2(2-x)^5 d x= $$

If $f(x)=\max \left\{x^3-4, x^4-4\right\}$ and $g(x)=\min \left\{x^2, x^3\right\}$, then $\int_{-1}^1(f(x)-g(x)) d x=$

$$ \int_0^1 \frac{2 x+5}{x^2+3 x+2} d x= $$

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

$$ \lim _{n \rightarrow \infty}\left[\begin{array}{c} \frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \\ \frac{9}{n^2}+\ldots+\frac{1}{n^2} \sec ^2 1 \end{array}\right]= $$

$$ \int_0^\pi\left(\sin ^5 x \cos ^3 x+\sin ^4 x \cos ^4 x+\sin ^3 x \cos ^4 x\right) d x= $$

$$ \int_0^1 \frac{x^4+1}{x^6+1} d x= $$

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

If $f(t)=\int_0^t \tan ^{(2 n-1)} x d x, n \in N$, then $f(t+\pi)=$

$$ \int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{\frac{5}{2}} d x= $$

$$ \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x= $$

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

$$ \int_0^{400 \pi} \sqrt{1-\cos 2 x} d x= $$

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

$$ \int_{\frac{5}{6}}^\pi \cos ^{-4} x d x= $$

$$ \int\limits_0^{\frac{3 \pi}{2}} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x= $$

If $k \in N$, then $\lim\limits_{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots .+\frac{1}{k n}\right]=$

$$ \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} d x= $$

$$ \int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x= $$

$$ \int_{5 \pi}^{25 \pi}|\sin 2 x+\cos 2 x| d x= $$

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

$$ \int_0^\pi \frac{x \sin x}{\sin ^2 x+2 \cos ^2 x} d x= $$

$$ \mathop {\lim }\limits_{n \to \infty }\left(\frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\ldots+\frac{n}{n^2+n^2}\right)= $$

$$ \int_0^{\frac{\pi}{2}} \log |\tan x+\cot x| d x= $$

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

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

Let $H(x)=3 x^4+6 x^3-2 x^2+1$ and $g(x)$ be a linear polynomial. If $\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x) +\frac{g(x)}{(x-1)(x+1)(x-2)}$, then

$H(-1)+2 H(2)-3 H(1)=$

$$ \int_{\pi / 4}^{\pi / 3} \frac{\cos x-\sin x}{\sin 2 x} d x= $$

$$ \int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x= $$

$$ \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x= $$

If $\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x=K \int_0^\pi \sin ^2 x d x+L \int_0^{\frac{\pi}{2}} \cos ^2 x d x$ and $K, L \in N$, then the number of possible ordered pairs ( $K, L$ ) is

$$ \mathop {\lim }\limits_{x \to \infty }\left[\left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\frac{27}{n^3}\right)^{\frac{9}{n^3}} \ldots . .(2)^{\frac{1}{n}}\right] \text { is equaln } $$

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

$$\int\limits_\pi ^\pi {}\frac{x \sin x}{1+\cos ^2 x} d x= $$

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

$$ \text { If } f(x)=\left\{\begin{array}{cc} \frac{6 x^2+1}{4 x^3+2 x+3} & , 0 < x < 1 \\ x^2+1 & , 1 \leq x < 2 \end{array} \text {, then } \int_0^2 f(x) d x=\right. $$

$$ \text { } \int\limits_{-3}^3|2-x| d x= $$

$$ \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{1}{\sqrt[5]{x^{30}+x^{25}}} d x= $$

Let $$T>0$$ be a fixed number. $$f: R \rightarrow R$$ is a continuous function such that $$f(x+T)=f(x), x \in R$$ If $$I=\int_\limits0^T f(x) d x$$, then $$\int_\limits0^{5 T} f(2 x) d x=$$

$$\int_\limits1^3 x^n \sqrt{x^2-1} d x=6 \text {, then } n=$$

[ . ] represents greatest integer function, then $$\int_{-1}^1(x[1+\sin \pi x]+1) d x=$$

$$\begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2 n+1}}+\frac{n}{(n+2) \sqrt{2(2 n+2)}}\right. \\ & \left.+\frac{n}{(n+3) \sqrt{3(2 n+3)}}+\ldots n \text { terms }\right]=\int_\limits0^1 f(x) d x \end{aligned}$$

then $$f(x)=$$

If $$I_n=\int_0^{\pi / 4} \tan ^n x d x$$, then $$\frac{1}{I_2+I_4}+\frac{1}{I_3+I_5}+\frac{1}{I_4+I_6}=$$

$$\int_0^{\pi / 4} e^{\tan ^2 \theta} \sin ^2 \theta \tan \theta d \theta=$$

$$\int_{\pi / 4}^{5 \pi / 4}(|\cos t| \sin t+|\sin t| \cos t) d t=$$

If $$f(x)=\max \{\sin x, \cos x\}$$ and $$g(x)=\min \{\sin x, \cos x\}$$, then $$\int_0^\pi f(x) d x+\int_0^\pi g(x) d x=$$

$$\int_0^1 a^k x^k d x=$$

Let $$\alpha$$ and $$\beta(\alpha<\beta)$$ are roots of $$18 x^2-9 \pi x+\pi^2=0, f(x)=x^2, g(x)=\cos x$$. Then, $$\int_\alpha^\beta x(g \circ f(x)) d x=$$

$$\int_0^\pi x\left(\sin ^2(\sin x)+\cos ^2(\cos x)\right) d x=$$

If $$\int_\limits0^\pi \log (\sin x) d x=8 k$$, then $$\int_\limits0^{\frac{\pi}{4}} \log (1+\tan x) d x$$ is equal to

If $$\int_\limits0^1 x^m(1-x)^n d x=k \int_\limits0^1 x^n(1-x)^m d x$$, then the value of $k$ equals

If $$\int_0^a {{{dx} \over {4 + {x^2}}} = {\pi \over 8}} $$, then the value of a is equal to

$$\int_1^2 {{{{x^3} - 1} \over {{x^2}}}} $$ is equal to

If $$\int_0^{\pi / 2} \tan ^n(x) d x=k \int_0^{\pi / 2} \cot ^n(x) d x$$, then

$$\int_0^2 x e^x d x$$ is equal to

$$\int_2^4\{|x-2|+|x-3|\} d x$$ is equal to

$$\int\limits_{-1 / 2}^{1 / 2}\left\{[x]+\log \left(\frac{1+x}{1-x}\right)\right\} d x$$ is equal to