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 \tan x \sec^2 x \sqrt{1 - \tan^2 x} \; dx = $

$\int \frac{1}{x^2 \sqrt{1 + x^2}} \, dx = $

Find $\int \cos 6x \sqrt{1+\sin 6x} \, dx$

$\int \tan^3 2x \sec 2x \, dx = $

The value of the integral $\int \frac{\sin(\ln(2 + 2x))}{x + 1} dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module