JEE Main 2019 — Mathematics Calculus

$$\Rightarrow I=\int x\sqrt{\frac{2sin({x}^{2}-1)-sin2({x}^{2}-1)}{2sin({x}^{2}-1)+sin2({x}^{2}-1)}}dx$$⇒I=∫x2sin(x2−1)+sin2(x2−1)2sin(x2−1)−sin2(x2−1)​​dx

Let, ${x}^{2}-1=\theta \Rightarrow xdx=\frac{1}{2}d\theta$x2−1=θ⇒xdx=21​dθ

$$\Rightarrow I=\frac{1}{2}\int \sqrt{\frac{2sin\theta -sin2\theta }{2sin\theta +sin2\theta }}d\theta$$⇒I=21​∫2sinθ+sin2θ2sinθ−sin2θ​​dθ

$$=\frac{1}{2}\int \sqrt{\frac{2sin\theta -2sin\theta cos\theta }{2sin\theta +2sin\theta cos\theta }}d\theta$$=21​∫2sinθ+2sinθcosθ2sinθ−2sinθcosθ​​dθ

$$=\frac{1}{2}\int \sqrt{\frac{1-cos\theta }{1+cos\theta }}d\theta$$=21​∫1+cosθ1−cosθ​​dθ

$$=\frac{1}{2}\int tan\frac{\theta }{2}d\theta$$=21​∫tan2θ​dθ

$=\frac{1}{2}\frac{{\mathrm{log}}_{e}(\mathrm{sec}\frac{\theta }{2})}{(\frac{1}{2})}+c$=21​(21​)loge​(sec2θ​)​+c, where $c$c is the constant of integration.

$$={\mathrm{log}}_{e}|{\mathrm{sec}}^{2}(\frac{{x}^{2}-1}{2})|+c$$=loge​∣sec2(2x2−1​)∣+c