Question

Evaluate: $\int\limits_{0}^{1}\frac{x \,tan^{-1}}{\left(1+x^{2}\right)^{3 /2}} dx$

• A. $\frac{4-\pi}{2\sqrt{2}}$
• B. $\frac{4+\pi}{2\sqrt{2}}$
• C. $\frac{4-\pi}{4\sqrt{2}}$
• D. None of these

Answer 1

Answer: (C)

Let $tan^{-1}x =\theta \Rightarrow x =tan \,\theta \Rightarrow dx = sec^{2}\,\theta \,d\theta $

Now, $x=0 \Rightarrow \theta = 0 \,and \, x=1 \Rightarrow \theta =\frac{\pi}{4} $

$\therefore I= \int\limits_{0}^{1} \frac{x\,tan^{-1}x}{\left(1+x^{2}\right)^{\frac{3}{2}}}dx = \int\limits_{0}^{\frac{\pi}{4}}\frac{ \theta \,tan\,\theta }{sec^{3}\,\theta } sec^{2}\,\theta \,d\theta$

$= \int\limits_{0}^{\frac{\pi }{4}}\theta \,sin\,\theta d\theta =\left[-\theta \, cos\,\theta \right]_{0}^{\frac{\pi}{4}}-\int\limits_{0}^{\frac{\pi}{4}}\left(-cos\,\theta \right)d\theta$

$=\left[-\theta \,cos\,\theta \right]_{0}^{\frac{\pi}{4}} +\left[sin\,\theta \right]_{0}^{\frac{\pi}{4}} = \frac{4-\pi}{4\sqrt{2}} $