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← JEE Main 2016 · 9 Apr · Shift 1
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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 75

If 2∫01tan⁡−1x dx=∫01cot⁡−1(1−x+x2) dx2\displaystyle\int_0^1\tan^{-1}x\,dx=\int_0^1\cot^{-1}(1-x+x^2)\,dx, then ∫01tan⁡−1(1−x+x2) dx\displaystyle\int_0^1\tan^{-1}(1-x+x^2)\,dx is equal to :

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Answer and worked solution

Official answer: C

∫01cot⁡−1(1−x+x2) dx=2∫01tan⁡−1x dx=π2−log⁡2\int_0^1\cot^{-1}(1-x+x^2)\,dx=2\int_0^1\tan^{-1}x\,dx=\frac\pi2-\log2. Since tan⁡−1u=π2−cot⁡−1u\tan^{-1}u=\frac\pi2-\cot^{-1}u, the required integral is π2−(π2−log⁡2)=log⁡2\frac\pi2-\left(\frac\pi2-\log2\right)=\log2.

Topics:Properties of definite integralsInverse trigonometric functions
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