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

Question

For $x^{2} 
eq n \pi+1, n \in N$ (the set of natural numbers), the integral $\int x \sqrt{\frac{2 \sin \left(x^{2}-1ight)-\sin 2\left(x^{2}-1ight)}{2 \sin \left(x^{2}-1ight)+\sin 2\left(x^{2}-1ight)}} d x$ is equal to (where $c$ is a constant of integration)

Professor PC Thomas Classes · Answer

(a, d) : Let\ $I=\int x \sqrt{\frac{2 \sin \left(x^{2}-1ight)-\sin 2\left(x^{2}-1ight)}{2 \sin \left(x^{2}-1ight)+\sin 2\left(x^{2}-1ight)}} d x$\ Put $x^{2}-1=t$\ $\Rightarrow \quad 2 x d x=d t$\ $	herefore \quad I=\frac{1}{2} \int \sqrt{\frac{2 \sin t-\sin 2 t}{2 \sin t+\sin 2 t}} d t$\ $=\frac{1}{2} \int \sqrt{\frac{1-\cos t}{1+\cos t}} d t$\ $=\frac{1}{2} \int 	an \frac{t}{2} d t=\log _{e}\left|\sec \frac{t}{2}ight|+c$\ $=\log _{e}\left|\sec \left(\frac{x^{2}-1}{2}ight)ight|+c$\ $=\frac{1}{2} \log _{e}\left|\sec ^{2}\left(\frac{x^{2}-1}{2}ight)ight|+c$\