Mathematics

Prove that : $$\displaystyle \int_{0}^{1} \tan^{-1} x dx = \dfrac {\pi}{4} - \dfrac {1}{2}\log 2$$.


SOLUTION
Suppose that
$$I = \displaystyle \int _{0}^{1}\tan^{-1} x dx .... (1)$$
By putting $$x = \tan \theta$$
$$\therefore dx = \sec^{2} \theta . d\theta$$
When $$x = 0$$ then $$\theta = 0$$
When $$x = 1$$ then $$\theta = \dfrac {\pi}{4}$$
$$\therefore$$ By equation $$(1)$$
$$I = \displaystyle \int _{0}^{\tfrac {\pi}{4}} \tan^{-1} (\tan \theta)\sec^{2}\theta . d\theta$$
$$= \displaystyle \int _{0}^{\tfrac {\pi}{4}}\theta . \sec^{2} \theta. d\theta$$
$$= [\theta . \displaystyle \int  \sec^{2} \theta d\theta]_{0}^{\tfrac {\pi}{4}} - \displaystyle \int _{0}^{\tfrac {\pi}{4}} \left [\dfrac {d}{d\theta} . \theta \displaystyle \int  \sec^{2} \theta d\theta \right ]d\theta$$
$$= [\theta . \tan \theta]_{0}^{\tfrac {\pi}{4}} - \displaystyle \int _{0}^{\tfrac {\pi}{4}} \tan \theta d\theta$$
$$= [\theta . \tan \theta ]_{0}^{\tfrac {\pi}{4}} + [\log \cos \theta ]_{0}^{\tfrac {\pi}{4}}$$
$$= \dfrac {\pi}{4},\tan \dfrac {\pi}{4} - 0 + \log \cos \dfrac {\pi}{4} - \log \cos 0$$
$$= \dfrac {\pi}{4}\times 1 + \log \dfrac {1}{\sqrt {2}} - \log (1)$$
$$= \dfrac {\pi}{4} + \log (1) - \log \sqrt {2} - \log (1)$$
$$= \dfrac {\pi}{4} - \log (2)^{\dfrac {1}{2}}$$
$$= \dfrac {\pi}{4} - \dfrac {1}{2}\log 2$$.
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Subjective Hard Published on 17th 09, 2020
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