Calculate the Integral again

Calculus Level 3

∫ − 1 1 sin ⁡ − 1 ( x ) cos ⁡ − 1 ( x ) d x = a − π 2 b \int_{-1}^1 \sin ^{-1}(x) \cos ^{-1}(x) \, dx=a-\frac{\pi ^2}{b}

where a , b a,b are positive integers. Submit a + b a+b .


The answer is 6.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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1 solution

Mark Hennings
Dec 16, 2017

With a couple of applications of Integration by Parts, we see that ∫ − 1 1 ( sin ⁡ − 1 x ) 2 d x = [ x ( sin ⁡ − 1 x ) 2 ] − 1 1 − 2 ∫ − 1 1 x sin ⁡ − 1 x 1 − x 2 d x = 1 2 π 2 − 2 [ − sin ⁡ − 1 x 1 − x 2 ] − 1 1 − 2 ∫ − 1 1 d x = 1 2 π 2 − 4 \begin{aligned} \int_{-1}^1 \big(\sin^{-1} x\big)^2\,dx & = \; \Big[x\big(\sin^{-1}x\big)^2\Big]_{-1}^1 - 2\int_{-1}^1 \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx \\ & = \; \tfrac12\pi^2 - 2\Big[-\sin^{-1}x\sqrt{1-x^2}\Big]_{-1}^1 - 2\int_{-1}^1\,dx \; = \; \tfrac12\pi^2 - 4 \end{aligned} Thus ∫ − 1 1 sin ⁡ − 1 x cos ⁡ − 1 x d x = ∫ − 1 1 sin ⁡ − 1 x ( 1 2 π − sin ⁡ − 1 x ) d x = 1 2 π ∫ − 1 1 sin ⁡ − 1 x d x − ∫ − 1 1 ( sin ⁡ − 1 x ) 2 d x = 4 − 1 2 π 2 \int_{-1}^1 \sin^{-1}x \cos^{-1}x\,dx \; = \; \int_{-1}^1 \sin^{-1}x\big(\tfrac12\pi - \sin^{-1}x\big)\,dx \; = \; \tfrac12\pi\int_{-1}^1 \sin^{-1}x\,dx - \int_{-1}^1 \big(\sin^{-1}x\big)^2\,dx \; = \; 4 - \tfrac12\pi^2 making the answer 4 + 2 = 6 4+2 = \boxed{6} .

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