Mathematics

Evaluate the definite integral   $$\displaystyle \int_0^{\frac {\pi}{2}}\sin 2x \tan^{-1}(\sin x)dx$$


SOLUTION
Let $$I\displaystyle =\int_0^{\frac {\pi}{2}}\sin 2x \tan^{-1}(\sin x)dx=\int_0^{\frac {\pi}{2}}2 \sin x \cos x \tan^{-1}(\sin x)dx$$
Also, let $$\sin x=t\Rightarrow \cos x dx=dt$$
When $$x=0, t=0$$ and when $$x=\dfrac {\pi}{2}, t=1$$
$$\Rightarrow\displaystyle I=2\int_0^1\tan^{-1}(t)dt$$ ............ (1)
Consider $$\displaystyle\int t\cdot \tan^{-1}t dt=\tan^{-1}t\cdot \int t dt-\int \left \{\frac {d}{dt}(\tan^{-1}t)\int t dt\right \}dt$$
$$\displaystyle=\tan^{-1}t\cdot \frac {t^2}{2}-\int \frac {1}{1+t^2}\cdot \frac {t^2}{2}dt$$
$$\displaystyle=\frac {t^2\tan^{-1}t}{2}-\frac {1}{2}\int \frac {t^2+1-1}{1+t^2}dt$$
$$\displaystyle=\frac {t^2\tan^{-1}t}{2}-\frac {1}{2}\int 1 dt+\frac {1}{2}\int \frac {1}{1+t^2}dt$$
$$\displaystyle=\frac {t^2\tan^{-1}t}{2}-\frac {1}{2}\cdot t+\frac {1}{2}\tan^{-1}t$$
$$\Rightarrow\displaystyle \int_0^1t\cdot \tan^{-1}t dt=\left [\frac {t^2\cdot\tan^{-1}t}{2}-\frac {t}{2}+\frac {1}{2}\tan^{-1}t\right ]_0^1$$
$$\displaystyle=\frac {1}{2}\left [\frac {\pi}{4}-1+\frac {\pi}{4}\right ]$$
$$\displaystyle=\frac {1}{2}\left [\frac {\pi}{2}-1\right ]=\frac {\pi}{4}-\frac {1}{2}$$
From equation (1), we obtain
$$I\displaystyle=2\left [\frac {\pi}{4}-\frac {1}{2}\right ]=\frac {\pi}{2}-1$$
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Subjective Medium Published on 17th 09, 2020
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