1 ∫ 0 X Tan − 1 X D X

Question

\[\int\limits_0^1 x \tan^{- 1} x\ dx\]

Solution

\[Let\ I = \int_0^1 x \tan^{- 1} x\ d\ x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \int_0^1 \frac{x^2}{1 + x^2} dx\]
\[ \Rightarrow I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \int_0^1 \left( \frac{1 + x^2}{1 + x^2} - \frac{1}{1 + x^2} \right) dx\]
\[ \Rightarrow I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \left[ x - \tan^{- 1} x \right]_0^1 \]
\[ \Rightarrow I = \frac{\pi}{8} - 0 - \frac{1}{2}\left( 1 - \frac{\pi}{4} - 0 \right)\]
\[ \Rightarrow I = \frac{\pi}{4} - \frac{1}{2}\]

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Definite Integrals

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Q 32 Q 31 Q 33

Chapter 19: Definite Integrals - Exercise 20.2 [Page 39]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12

Chapter 19 Definite Integrals
Exercise 20.2 | Q 32 | Page 39

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