Question

Evaluate: $\underset{0}{\overset{1}{\int }}\frac{xta{n}^{-1}}{{\left(1+x^2\right)}^{3/2}}dx$

• 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 $ta{n}^{-1}x=\theta \Rightarrow x=tan\theta \Rightarrow dx=se{c}^2\theta d\theta$

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

$\therefore I=\underset{0}{\overset{1}{\int }}\frac{xta{n}^{-1}x}{{\left(1+x^2\right)}^{\frac{3}{2}}}dx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\theta tan\theta }{se{c}^3\theta }se{c}^2\theta d\theta$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\theta sin\theta d\theta ={\left[-\theta cos\theta \right]}_0^{\frac{\pi }{4}}-\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(-cos\theta \right)d\theta$

$={\left[-\theta cos\theta \right]}_0^{\frac{\pi }{4}}+{\left[sin\theta \right]}_0^{\frac{\pi }{4}}=\frac{4-\pi }{4\sqrt{2}}$