Prove the following

Question:

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

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

is equal to :

(where $\mathrm{c}$ is a constant of integration)

  1. $\log _{\mathrm{e}}\left|\sec \left(\frac{\mathrm{x}^{2}-1}{2}\right)\right|+\mathrm{c}$

  2. $\log _{e}\left|\frac{1}{2} \sec ^{2}\left(x^{2}-1\right)\right|+c$

  3. $\frac{1}{2} \log _{e}\left|\sec ^{2}\left(\frac{\mathrm{x}^{2}-1}{2}\right)\right|+\mathrm{c}$

  4. $\frac{1}{2} \log _{e}\left|\sec \left(x^{2}-1\right)\right|+c$


Correct Option: 1

Solution:

Put $\left(x^{2}-1\right)=1$

$\Rightarrow 2 x d x=d t$

$\therefore \quad \mathrm{I}=\frac{1}{2} \int \sqrt{\frac{1-\cos t}{1+\cos t}} d t$

$=\frac{1}{2} \int \tan \left(\frac{\mathrm{t}}{2}\right) \mathrm{dt}$

$=\ln \left|\sec \frac{\mathrm{t}}{2}\right|+\mathrm{c}$

$I=\ln \left|\sec \left(\frac{x^{2}-1}{2}\right)\right|+c$

Leave a comment