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JEE Previous Year Question

$\int_{0}^{1} x tan^{- 1} xdx =$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $\frac{π}{4} + \frac{1}{2}$
B $\frac{π}{4} - \frac{1}{2}$
C $\frac{1}{2} - \frac{π}{4}$
D $- \frac{π}{4} - \frac{1}{2}$

Correct answer: B

Solution

(b) : We have, \int_{0}^{1} x \cdot tan^{- 1} xdx = (tan^{- 1} x \cdot \frac{x ^{2}}{2})_{0}^{1} - \int_{0}^{1} \frac{1}{1 + x ^{2}} \cdot \frac{x ^{2}}{2} dx = (\frac{π}{4} \cdot \frac{1}{2} - 0) - \frac{1}{2} \int_{0}^{1} \frac{1 + x ^{2} - 1}{1 + x ^{2}} dx = \frac{π}{8} - \frac{1}{2} \int_{0}^{1} (1 - \frac{1}{1 + x ^{2}}) dx = \frac{π}{8} - \frac{1}{2} (x - tan^{- 1} x)_{0}^{1} = \frac{π}{8} - \frac{1}{2} ((1 - \frac{π}{4}) - (0)) = \frac{π}{8} - \frac{1}{2} + \frac{π}{8} = \frac{π}{4} - \frac{1}{2}

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