Embibe Experts Solutions for Chapter: Definite Integration, Exercise 1: Exercise 1

Evaluate :  $\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{1+{\cos }^{2}x}dx$

If  $f\left(x\right)={\int }_{x^2}^{x^2+1}{e}^{-t^2}dt,$  then  $f\left(x\right)$  increases in:

If $f\left(x\right)={\int }_1^x\left(\frac{\ln t}{1+t}\right)dt$, where $n>0$ then the value $\left(s\right)$ of $x$ satisfying the equation, $f\left(x\right)+f\left(\frac{1}{x}\right)=2$ is-

The value of integral ${\int }_0^{\mathrm{\pi }}xf\left(\sin x\right)dx=$

The function $f\left(x\right)={\int }_0^x\sqrt{1-t^4}dt$ is such that :

Let $S_n=\frac{n}{\left(n+1\right)\left(n+2\right)}+\frac{n}{\left(n+2\right)\left(n+4\right)}+\frac{n}{\left(n+3\right)\left(n+6\right)}+\dots \dots ..+\frac{1}{6n}$, then $\underset{n\to \infty }{\lim }{S}_n$ is

The value of ${\int }_0^1\left(\frac{1-x}{1+x}\right)\frac{dx}{\sqrt{x+x^2+x^3}}$ is equal to

If ${\int }_1^2\frac{\left(x^2-1\right)dx}{x^3·\sqrt{2x^4-2x^2+1}}=\frac{u}{v}$ where $u$ and $v$ are in their lowest form, then the value of $\frac{\left(1000\right)u}{v}$ is