Question

For $x^2\ne n\pi +1,n\in N$ (the set of natural numbers), the integral

$\int x\sqrt{\frac{2\sin (x^2-1)-\sin 2(x^2-1)}{2\sin (x^2-1)+\sin 2(x^2-1)}}dx$ is equal to :
(where $C$ is a constant of integration)

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

Answer 1

The correct option is B ${\log }_e\left|\sec \left(\frac{x^2-1}{2}\right)\right|+C$

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

Putting $x^2-1=\theta \Rightarrow 2xdx=d\theta$ then,
$I=\int x\sqrt{\frac{2\sin (x^2-1)-\sin 2(x^2-1)}{2\sin (x^2-1)+\sin 2(x^2-1)}}dx=\frac{1}{2}\int \sqrt{\frac{2\sin \theta -\sin 2\theta }{2\sin \theta +\sin 2\theta }}d\theta =\frac{1}{2}\int \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}d\theta =\frac{1}{2}\int \sqrt{\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}}d\theta =\frac{1}{2}\int \sqrt{{\tan }^2\frac{\theta }{2}}d\theta =\frac{1}{2}\int \tan \frac{\theta }{2}d\theta ={\log }_e\left|{\sec }^2\frac{\theta }{2}\right|+C$
where $C$ is the constant of integration.

Hence the value of the integration,
$I={\log }_e\left|{\sec }^2\frac{(x^2-1)}{2}\right|+C$