Question

Evaluate the definite integral

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

Answer 1

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

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

Let $\sin x=t$

Differentiating both sides w.r.t $x$

$\cos x=\frac{dt}{dx}$

$\Rightarrow dx=\frac{dt}{\cos x}$

When $x=0\Rightarrow t=0$

When $x=\frac{\pi }{2}\Rightarrow t=1$

Substituting $x$ and $dx$

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

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

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

We know that $\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int (\frac{d}{dx}\left(f\left(x\right)\right)\int g\left(x\right))dx$

Putting $f\left(x\right)={\tan }^{-1}t$ and $g\left(x\right)=t$

$=2({\tan }^{-1}t\int tdt-\int \left(\frac{d}{dt}{\tan }^{-1}t\right)\int tdtdt)$

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

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

$=t^2{\tan }^{-1}t-\int \frac{t^2}{1+t^2}dt$

Let $I_1=\int \frac{t^{2}dt}{1+t^2}$

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

$=\int dt-\int \frac{dt}{1+t^2}$

$=t-{\tan }^{-1}t$

Thus, our equation becomes

$\therefore \int {\tan }^{-1}t\times tdt=t^2{\tan }^{-1}t-I_1$

$=t^2{\tan }^{-1}t-(t-{\tan }^{-1}t)$

$=t^2{\tan }^{-1}t-t+{\tan }^{-1}t$

$=F\left(x\right)$

$2{\int }_0^{1}t\times {\tan }^{-1}tdt=F\left(1\right)-F\left(0\right)$

$=(1\times {\tan }^{-1}1-1+{\tan }^{-1}1)(-0+{\tan }^{-1}0)$

$=\frac{\pi }{4}-1+\frac{\pi }{4}-(0-0+0)$

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