for x


For $x^{2} \neq \mathrm{n} \pi+1, \mathrm{n} \in \mathrm{N}$ (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 $c$ is a constant of integration)

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

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

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

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

Correct Option: , 3


Consider the given integral

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

$(\therefore \sin 2 \theta=2 \sin \theta \cos \theta)$

$\Rightarrow \quad I=\int x \sqrt{\frac{1-\cos \left(x^{2}-1\right)}{1+\cos \left(x^{2}-1\right)}} d x$

$\Rightarrow I=\int x\left|\tan \left(\frac{x^{2}-1}{2}\right)\right| d x$

Now let $\frac{x^{2}-1}{2}=t \quad \Rightarrow \quad \frac{2 x}{2} d x=d t$

$\therefore \quad I=\int|\tan (t)| d t=\ln |\sec t|+C$

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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now