Question

${\int }_0^{\pi /2}sin2xta{n}^{-1}\left(sinx\right)dx=$

• A. $\frac{\pi }{2}$-1
  
• B. $\frac{\pi }{2}$+1
• C. $\frac{3\pi }{2}$+1
• D. $\frac{3\pi }{2}$-1

Answer 1

The correct option is A $\frac{\pi }{2}$-1

We have,

$I={\int }_0^{\frac{\pi }{2}}\sin 2x{\tan }^{-1}\left(\sin x\right)dx$

$={\int }_0^{\frac{\pi }{2}}2\sin x\cos x{\tan }^{-1}\left(\sin x\right)dx$

Let

$\sin x=t$

$\cos xdx=dt$

Change limit

$\sin 0=t$

$t=0$

And,

$\sin \frac{\pi }{2}=t$

$t=1$

Then,

${\int }_0^{1}2t{\tan }^{-1}t\cos xdx$

$={\int }_0^{1}2t{\tan }^{-1}tdt$

$=2{\int }_0^{1}t{\tan }^{-1}tdt$

On integrating and we get,

$2[{\tan }^{-1}t{\int }_0^{1}tdt-{\int }_0^1\left(\frac{d\left({\tan }^{-1}t\right)}{dt}{\int }_0^{1}tdt\right)dt]$

$=2[{\tan }^{-1}t\left[{{\left(\frac{t^2}{2}\right)}_0}^1\right]-{\int }_0^1\frac{1}{1+t^2}\frac{t^2}{2}dt]$

$=2{\tan }^{-1}t{{\left(\frac{t^2}{2}\right)}_0}^1-{\int }_0^1\frac{t^2}{1+t^2}dt$

$=2{\tan }^{-1}t{{\left(\frac{t^2}{2}\right)}_0}^1-{\int }_0^1\frac{t^2+1-1}{1+t^2}dt$

$=2{\tan }^{-1}t{{\left(\frac{t^2}{2}\right)}_0}^1-{\int }_0^1\frac{t^2+1}{1+t^2}dt+{\int }_0^1\frac{1}{1+t^2}dt$

$=2{\tan }^{-1}t{{\left(\frac{t^2}{2}\right)}_0}^1-{\int }_0^{1}1dt+{\int }_0^1\frac{1}{1+t^2}dt$

$=2{\tan }^{-1}t{{\left(\frac{t^2}{2}\right)}_0}^1-{{\left[t\right]}_0}^1+{{\left[{\tan }^{-1}t\right]}_0}^1+C$

$=2[{\tan }^{-1}1-{\tan }^{-1}0][\frac{1^2}{2}-\frac{0^2}{2}]-[1-0]+[{\tan }^{-1}1-{\tan }^{-1}0]+C$

$=2[\frac{\pi }{4}-0]\left[\frac{1}{2}\right]-1+[\frac{\pi }{4}-0]$

$=\frac{\pi }{4}+\frac{\pi }{4}-1$

$=\frac{2\pi }{4}-1$

$=\frac{\pi }{2}-1$

Hence, this is the answer.