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Calculus Level 2

∫ 0 1 tan ⁡ − 1 x 1 + x 2 d x = ? \large \int_{0}^{1} \frac{\tan^{-1} x}{1 + x^{2}} \mathrm dx = ?

π 2 32 \frac{\pi^{2}}{32} π 2 16 \frac{\pi^{2}}{16} π 2 4 \frac{\pi^{2}}{4} None of the others. π 2 8 \frac{\pi^{2}}{8}

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1 solution

Tapas Mazumdar
Apr 9, 2018

I = ∫ 0 1 tan ⁡ − 1 x 1 + x 2 d x Let tan ⁡ − 1 x = u ⟹ d x 1 + x 2 = d u and ∫ 0 1 → ∫ 0 π / 4 = ∫ 0 π / 4 u d u = 1 2 ( π 4 ) 2 = π 2 32 \begin{aligned} I &= \displaystyle \int_0^1 \dfrac{\tan^{-1} x}{1+x^2} \,dx & \small\color{#3D99F6}{\text{Let } \tan^{-1} x = u \implies \dfrac{dx}{1+x^2} = \,du \text{ and } \displaystyle \int_0^1 \rightarrow \int_0^{{\pi}/{4}} } \\ &= \displaystyle \int_0^{{\pi}/{4}} u \,du \\ &= \dfrac 12 {\left( \dfrac{\pi}{4} \right)}^2 \\ &= \boxed{\dfrac{\pi^2}{32}} \end{aligned}

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