Question

Evaluate the definite integral: $∫_0^{\frac{π}{2}}sin2x\,tan^{-1}(sinx)dx$

Answer 1

$Let I=∫_0^{\frac{π}{2}}=sin2xtan^{-1}(sinx)dx=∫_0^{\frac{π}{2}}2sinx\,cosx\,tan^{-1}(sinx)dx$

$Also,let\, sinx=t⇒cosxdx=dt$

$When\, x=0,t=0\, and\, when\, x=\frac{π}{2},t=1$

$⇒I=2∫_0^1t\,tan^{-1}(t)dt...(1)$

$Consider ∫t.tan^{-1}t\,dt=tan^{-1}t.∫t\,dt-∫[\frac{d}{dt}(tan^{-1}t)∫tdt]dt$

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

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

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

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

$⇒∫_0^1t.tan^{-1}t\,dt=\bigg[\frac{t^2.tan^{-1}t}{2}-\frac{t}{2}+\frac{1}{2}tan^{-1}t\bigg]_0^1$

$=\frac{1}{2}[\frac{π}{4}-1+\frac{\pi}{4}]$

$=\frac{1}{2}[\frac{π}{2}-1]=\frac{π}4-\frac{1}{2}$

From equation(1),we obtain

$I=2[\frac{π}{4}-\frac{1}{2}]=\frac{π}{2}-1$