Definite Integration · Mathematics · AP EAPCET

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

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

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

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

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

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

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

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 $$\int_0^a {{{dx} \over {4 + {x^2}}} = {\pi \over 8}} $$, 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