Evaluate: $\int_0^1 \sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)dx$

Evaluate: ∫₀¹ sin⁻¹((2x)/(1+x²))dx

Miscellaneous Exercise 4 · Q75

\int_0^1 \sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)dx

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As in III.1, $x=\tan\theta$ turns $\sin^{-1}(2x/(1+x^2))$ into $2\theta$ , but here $dx=\sec^2\theta\,d\theta$ (no cancelling weight this time).

\int_0^{\pi/4}2\theta\sec^2\theta\,d\theta.

IBP with $u=2\theta,\ dv=\sec^2\theta\,d\theta\Rightarrow v=\tan\theta$ :

$2\theta\tan\theta-\int2\tan\theta\,d\theta=2\theta\tan\theta+2\ln|\cos\theta|.$ …