Will the ILATE rule work?

Calculus Level 5

∫ − 1 3 1 3 x 4 1 − x 4 cos ⁡ − 1 ( 2 x 1 + x 2 ) d x \large \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{x^{4}}{1-x^{4}}\cos^{-1}\left(\frac{2x}{1+x^{2}}\right)dx The closed form of the integral above is of the form π 2 a − π b − π c ln ⁡ ( b − 1 b + 1 ) \displaystyle \frac{\pi ^{2}}{a}-\frac{\pi }{\sqrt{b}}-\frac{\pi }{c}\ln \left(\frac{\sqrt{b}-1}{\sqrt{b}+1}\right) .

Find the value of a + b + c a+b+c .


The answer is 19.

Aaron Jerry Ninan by Aaron Jerry Ninan , 20, India

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1 solution

Mark Hennings
Feb 5, 2017

With the substitution x = tan ⁡ 1 2 θ x = \tan\tfrac12\theta we have I = ∫ − 1 3 1 3 x 4 1 − x 4 cos ⁡ − 1 ( 2 x 1 + x 2 ) d x = ∫ − 1 3 1 3 x 4 1 − x 2 cos ⁡ − 1 ( 2 x 1 + x 2 ) d x 1 + x 2 = ∫ − 1 3 1 3 ( 1 1 − x 2 − x 2 − 1 ) cos ⁡ − 1 ( 2 x 1 + x 2 ) d x 1 + x 2 = 1 2 ∫ − 1 3 π 1 3 π ( cos ⁡ 2 1 2 θ cos ⁡ θ − sec ⁡ 2 1 2 θ ) cos ⁡ − 1 ( sin ⁡ θ ) d θ = 1 2 ∫ − 1 3 π 1 3 π ( 1 2 + 1 2 sec ⁡ θ − sec ⁡ 2 1 2 θ ) ( 1 2 π − θ ) d θ \begin{aligned} I \; = \; \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{x^4}{1-x^4} \cos^{-1}\left(\frac{2x}{1+x^2}\right)\,dx & = \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{x^4}{1-x^2} \cos^{-1}\left(\frac{2x}{1+x^2}\right)\,\frac{dx}{1+x^2} \\ & = \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \left(\frac{1}{1-x^2} - x^2 - 1\right) \cos^{-1}\left(\frac{2x}{1+x^2}\right)\,\frac{dx}{1+x^2} \\ & = \tfrac12\int_{-\frac13\pi}^{\frac13\pi}\left(\frac{\cos^2\tfrac12\theta}{\cos\theta} - \sec^2\tfrac12\theta\right) \cos^{-1}(\sin\theta)\,d\theta \\ & = \tfrac12\int_{-\frac13\pi}^{\frac13\pi} \left(\tfrac12 + \tfrac12\sec\theta - \sec^2\tfrac12\theta\right)\big(\tfrac12\pi - \theta\big)\,d\theta \end{aligned} By symmetry we see that I = 1 4 π ∫ − 1 3 π 1 3 π ( 1 2 + 1 2 sec ⁡ θ − sec ⁡ 2 1 2 θ ) d θ = 1 4 π [ 1 2 θ + 1 2 ln ⁡ ( sec ⁡ θ + tan ⁡ θ ) − 2 tan ⁡ 1 2 θ ] − 1 3 π 1 3 π = 1 4 π [ ( 1 6 π + 1 2 ln ⁡ ( 2 + 3 ) − 2 3 ) − ( − 1 6 π + 1 2 ln ⁡ ( 2 − 3 ) + 2 3 ) ] = 1 4 π [ 1 3 π − 4 3 + 1 2 ln ⁡ ( 2 + 3 2 − 3 ) ] = 1 12 π 2 − π 3 − 1 4 π ln ⁡ ( 3 − 1 3 + 1 ) \begin{aligned} I & = \tfrac14\pi\int_{-\frac13\pi}^{\frac13\pi}\left(\tfrac12 + \tfrac12\sec\theta - \sec^2\tfrac12\theta\right)\,d\theta \\ & = \tfrac14\pi \Big[\tfrac12\theta + \tfrac12\ln(\sec\theta + \tan\theta) - 2\tan\tfrac12\theta \Big]_{-\frac13\pi}^{\frac13\pi} \\ & = \tfrac14\pi \Big[\big(\tfrac16\pi + \tfrac12\ln(2 + \sqrt{3}) - \tfrac{2}{\sqrt{3}}\big) - \big(-\tfrac16\pi + \tfrac12\ln(2 - \sqrt{3}) + \tfrac{2}{\sqrt{3}}\big)\Big] \\ & = \tfrac14\pi \Big[\tfrac13\pi - \tfrac{4}{\sqrt{3}} + \tfrac12\ln \left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)\Big] \\ & = \tfrac{1}{12}\pi^2 - \tfrac{\pi}{\sqrt{3}} - \tfrac14\pi\ln\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) \end{aligned} making the answer 12 + 3 + 4 = 19 12 + 3 + 4 = \boxed{19} .

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Good approach!!. Another method to solve this problem is to replace the cos inverse term with (pi/2) - sin inverse of that. By expanding brackets we get two functions , one having the sin inverse term will get zero as it is an odd function, the other function can be easily evaluated using standard integrals.

Aaron Jerry Ninan - 4 years, 4 months ago

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You are applying symmetry (oddness) and then integrating. I substituted first. Not a lot of difference, since the tan ⁡ 1 2 θ \tan\tfrac12\theta substitution is definitely part of the standard integral armoury.

Mark Hennings - 4 years, 4 months ago

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