Question

$\int x.\sqrt{\left[\frac{2\sin (x^2-1)-\sin 2(x^2-1)}{2\sin (x^2-1)+\sin 2(x^2-1)}\right]}dx$ where $x^2-1\ne n\pi$

• A. $\log \sec x^2-1.$
• B. $\log \sec \frac{x^2-1}{2}.$
• C. $\log \tan \frac{x^2-1}{2}.$
• D. $\log \cos \frac{x^2-1}{2}.$

Answer 1

Answer: (B)

$\log \sec \frac{x^2-1}{2}.$

Let $I=\int x\sqrt{\left[\frac{2\sin (x^2-1)-\sin 2(x^2-1)}{2\sin (x^2-1)+\sin 2(x^2-1)}\right]}dx$
Put $x^2-1=t\Rightarrow 2xdx=dt$
Using
${\left(N^{\prime }\right)}^2\Rightarrow 2\sin t-\sin 2t=2\sin t(1-\cos t){\left(D^{\prime }\right)}^2\Rightarrow 2\sin t(1+\cos t)$
And $\frac{1-\cos t}{1+\cos t}=\frac{2{\sin }^2\left(t/2\right)}{2{\cos }^2\left(t/2\right)}={\tan }^2\frac{t}{2}$
Therefore
$I=\frac{1}{2}\int \tan \frac{t}{2}dt=\log \sec \frac{t}{2}=\log \sec \frac{x^2-1}{2}$
Hence, option 'B' is correct.