For ${x^2} \ne n\pi + 1, n \in N$ (the set of natural numbers),the integral $\int {x\sqrt {\frac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } dx$ is

  • A
    ${\log _e}\left| {\frac{1}{2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c$
  • B
    $\frac{1}{2}{\log _e}\left| {\sec \left( {{x^2} - 1} \right)} \right| + c$
  • C
    $\frac{1}{2}{\log _e}\left| {{{\sec }^2}\left( {\frac{{{x^2} - 1}}{2}} \right)} \right| + c$
  • D
    ${\log _e}\left| {\sec \left( {\frac{{{x^2} - 1}}{2}} \right)} \right| + c$

Explore More

Similar Questions

If $\int \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x=f(x)+C$,then $e^{f(x)}=$

If $\int \frac{\sqrt{1-x^2}}{x^4} \,dx = A(x)\left(\sqrt{1-x^2}\right)^{m} + c$ for a suitably chosen integer $m$ and a function $A(x)$,where $c$ is a constant of integration,then $(A(x))^{m}$ equals

$\int \frac{2x \tan^{-1}(x^2)}{1 + x^4} \, dx = $

Integrate the function: $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

Difficult
View Solution

$\int \frac{dx}{e^x+e^{-x}+2} = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo