int x sqrt{frac{2 sin left(x^2+1right)-sin 2l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} \, dx$ Using

Vedclass · Accepted Answer

$\log \left(\sec \left(\frac{x^2+1}{2}\right)\right)+c$,where $c$ is a constant of integration.

Vedclass · Answer

(A) Let $I = \int x \sqrt{\frac{2 \sin \left(x^2+1ight)-\sin 2\left(x^2+1ight)}{2 \sin \left(x^2+1ight)+\sin 2\left(x^2+1ight)}} \, dx$ Using $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get: $I = \int x \sqrt{\frac{2 \sin \left(x^2+1ight) - 2 \sin \left(x^2+1ight) \cos \left(x^2+1ight)}{2 \sin \left(x^2+1ight) + 2 \sin \left(x^2+1ight) \cos \left(x^2+1ight)}} \, dx$ $I = \int x \sqrt{\frac{1 - \cos \left(x^2+1ight)}{1 + \cos \left(x^2+1ight)}} \, dx$ Using $1 - \cos 	heta = 2 \sin^2(	heta/2)$ and $1 + \cos 	heta = 2 \cos^2(	heta/2)$: $I = \int x \sqrt{\frac{2 \sin^2 \left(\frac{x^2+1}{2}ight)}{2 \cos^2 \left(\frac{x^2+1}{2}ight)}} \, dx$ $I = \int x 	an \left(\frac{x^2+1}{2}ight) \, dx$ Let $t = \frac{x^2+1}{2}$,then $dt = x \, dx$. $I = \int 	an t \, dt = \log |\sec t| + c$ $I = \log \left| \sec \left(\frac{x^2+1}{2}ight) ight| + c$

$\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} \, dx =$

Similar Questions

$\int 3^{3^x} \cdot 3^x \, dx =$

Integrate the function $\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$.

If $\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$,then $g(x)$ is equal to

$\int \frac{1}{x+x \log x} d x=$ . . . . . . .

$\int \frac{\sin(2x)}{\sin^2(x) + 2\cos^2(x)} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module