Question

\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]

Answer 1

We have,

\[I = \int_0^1 x \left( \tan^{- 1} x \right)^2 d x\]

\[\text{Putting }\tan^{- 1} x = u\]

\[ \Rightarrow x = \tan u\]

\[ \Rightarrow dx = \sec^2 u du\]

\[\text{When }x \to 0; u \to 0\]

\[\text{and }x \to 1; u \to \frac{\pi}{4}\]

\[ \therefore I = \int_0^\frac{\pi}{4} \left( \tan u \right) u^2 \sec^2 u\ du\]

\[ = \int_0^\frac{\pi}{4} u^2 \tan u \sec^2 u\ du\]

\[ = \left[ u^2 \frac{\tan^2 u}{2} \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} 2u \frac{\tan^2 u}{2} du\]

\[ = \left[ u^2 \frac{\tan^2 u}{2} \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} u \left( \sec^2 u - 1 \right) du\]

\[ = \left[ u^2 \frac{\tan^2 u}{2} \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} u \sec^2 u\ du + \int_0^\frac{\pi}{4} u\ du\]

\[ = \left[ u^2 \frac{\tan^2 u}{2} \right]_0^\frac{\pi}{4} - \left[ u \tan u \right]_0^\frac{\pi}{4} + \int_0^\frac{\pi}{4} \tan u\ du + \left[ \frac{u^2}{2} \right]_0^\frac{\pi}{4} \]

\[ = \left[ u^2 \frac{\tan^2 u}{2} \right]_0^\frac{\pi}{4} - \left[ u \tan u \right]_0^\frac{\pi}{4} + \left[ \log \left| \sec u \right| \right]_0^\frac{\pi}{4} + \left[ \frac{u^2}{2} \right]_0^\frac{\pi}{4} \]

\[ = \frac{\pi^2}{16} \times \frac{1}{2} - \frac{\pi}{4} + \log\sqrt{2} + \frac{\pi^2}{32}\]

\[ = \frac{\pi^2}{16} - \frac{\pi}{4} + \log\sqrt{2}\]

\[ = \frac{\pi^2}{16} - \frac{\pi}{4} + \frac{1}{2}\log 2\]