Integrate it I guess

Calculus Level 3

∫ − 1 1 cos ⁡ − 1 ( x ) tan ⁡ − 1 ( x ) d x = − π 2 a − π sinh ⁡ − 1 ( b ) + 1 2 π log ⁡ ( c ) + π 2 − π \int_{-1}^1 \cos ^{-1}(x) \tan ^{-1}(x) \, dx=-\frac{\pi ^2}{a}-\pi \sinh ^{-1}(b)+\frac{1}{2} \pi \log (c)+\pi \sqrt{2}-\pi

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


The answer is 13.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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1 solution

Mark Hennings
Dec 16, 2017

We need the answer to this other question. Note that ∫ − 1 1 cos ⁡ − 1 x tan ⁡ − 1 x d x = ∫ − 1 1 ( 1 2 π − sin ⁡ − 1 x ) tan ⁡ − 1 x d x = − ∫ − 1 1 sin ⁡ − 1 x tan ⁡ − 1 x d x \int_{-1}^1 \cos^{-1}x \tan^{-1}x\,dx \; = \; \int_{-1}^1 \big(\tfrac12\pi - \sin^{-1}x\big)\tan^{-1}x\,dx \; = \; -\int_{-1}^1 \sin^{-1}x \tan^{-1}x\,dx Manipulating the result from the previous question, this integral is equal to − 1 4 π 2 − π sinh ⁡ − 1 1 + 1 2 π ln ⁡ 8 + π 2 − π -\tfrac14\pi^2 - \pi\sinh^{-1}1 + \tfrac12\pi\ln8 + \pi\sqrt{2} - \pi making the answer 4 + 1 + 8 = 13 4 + 1 + 8 = \boxed{13} .

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