Standard values

Calculus Level 4

Define the following integral

ξ ( θ ) = ∫ 0 1 sin ⁡ − 1 ( θ x ) cos ⁡ − 1 ( θ x ) d x \xi(\theta )=\int_{0}^{1}\sin^{-1}\,(\theta x)\cos^{-1}\,(\theta x)\,dx where 0 < θ ≤ 1 0 < \theta \leq 1 and

ξ ( 1 ) + ξ ( 1 2 ) + ξ ( 1 2 ) + ξ ( 3 2 ) = a + b c π 2 − d e π − f e + g d g g π \xi(1)+\xi\left(\frac{1}{2}\right)+\xi\left(\frac{1}{\sqrt 2}\right)+\xi\left(\frac{\sqrt 3}{2}\right)= a+\dfrac{b}{c}\pi^2-\dfrac{d}{e}\pi-\dfrac{f\sqrt e +g\sqrt d }{g\sqrt g}\pi

where a a , b b , c c , d d , e e , f f , and g g are positive integers with gcd ⁡ ( b , c ) = gcd ⁡ ( d , e ) = 1 \gcd(b,c) = \gcd(d,e) = 1 .

Find a + b + c + d + 3 e + f + g \sqrt{a+b+c+d+3e+f+g} .

Note : This is an original problem and generalized problem for Inverse Trig Integral .


The answer is 14.

Naren Bhandari by Naren Bhandari , 21, India

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2 solutions

Chew-Seong Cheong
Jan 28, 2019

ξ ( θ ) = ∫ 0 1 sin ⁡ − 1 ( θ x ) cos ⁡ − 1 ( θ x ) d x Let u = θ x ⟹ d u = θ d x = 1 θ ∫ 0 θ sin ⁡ − 1 u cos ⁡ − 1 u d u Note that cos ⁡ − 1 u = π 2 − sin ⁡ − 1 u = π 2 θ ∫ 0 θ sin ⁡ − 1 u d u − 1 θ ∫ 0 θ ( sin ⁡ − 1 u ) 2 d u By integration by parts \begin{aligned} \xi (\theta) & = \int_0^1 \sin^{-1} (\theta x) \cos^{-1} (\theta x) \ dx & \small \color{#3D99F6} \text{Let }u = \theta x \implies du = \theta \ dx \\ & = \frac 1\theta \int_0^\theta \sin^{-1} u \cos^{-1} u \ du & \small \color{#3D99F6} \text{Note that }\cos^{-1} u = \frac \pi 2 - \sin^{-1} u \\ & = \frac \pi{2\theta} \int_0^\theta \sin^{-1} u \ du - \frac 1\theta \int_0^\theta \left(\sin^{-1} u\right)^2 \ du & \small \color{#3D99F6} \text{By integration by parts} \end{aligned}

= π 2 θ [ u sin ⁡ − 1 u − ∫ u 1 − u 2 d u ] 0 θ − 1 θ [ u ( sin ⁡ − 1 u ) 2 − ∫ 2 u sin ⁡ − 1 u 1 − u 2 d u ] 0 θ = π 2 θ [ u sin ⁡ − 1 u + 1 − u 2 ] 0 θ − 1 θ [ u ( sin ⁡ − 1 u ) 2 + 2 1 − u 2 sin ⁡ − 1 u − ∫ 2 d u ] 0 θ = π 2 θ ( θ sin ⁡ − 1 θ + 1 − θ 2 − 1 ) − 1 θ ( θ ( sin ⁡ − 1 θ ) 2 + 2 1 − θ 2 sin ⁡ − 1 θ − 2 θ ) = 2 + sin ⁡ − 1 θ cos ⁡ − 1 θ − π 2 θ − 1 − θ 2 θ ( sin ⁡ − 1 θ − cos ⁡ − 1 θ ) \begin{aligned} \quad \ \ & = \frac \pi{2\theta} \left[u\sin^{-1} u - \int \frac u{\sqrt{1-u^2}} du \right]_0^\theta - \frac 1\theta \left[u \left(\sin^{-1} u\right)^2 - \int \frac {2u \sin^{-1}u}{\sqrt{1-u^2}} du \right]_0^\theta \\ & = \frac \pi{2\theta} \left[u\sin^{-1} u + \sqrt{1-u^2} \right]_0^\theta - \frac 1\theta \left[u \left(\sin^{-1} u\right)^2 + 2\sqrt{1-u^2} \sin^{-1}u - \int 2\ du \right]_0^\theta \\ & = \frac \pi{2\theta} \left(\theta \sin^{-1} \theta + \sqrt{1-\theta^2} - 1 \right) - \frac 1\theta \left(\theta \left(\sin^{-1} \theta \right)^2 + 2 \sqrt{1-\theta^2} \sin^{-1}\theta - 2\theta \right) \\ & = 2 + \sin^{-1} \theta \cos^{-1} \theta - \frac \pi{2\theta} - \frac {\sqrt{1-\theta^2}}\theta \left(\sin^{-1} \theta - \cos^{-1} \theta \right) \end{aligned}

Then, we have:

ξ ( 1 ) = 2 + π 2 × 0 − π 2 − 0 = 2 − π 2 ξ ( 1 2 ) = 2 + π 6 × π 3 − π − 3 ( π 6 − π 3 ) = 2 + π 2 18 − π + π 2 3 ξ ( 1 2 ) = 2 + π 4 × π 4 − π 2 − ( π 4 − π 4 ) = 2 + π 2 16 − π 2 ξ ( 3 2 ) = 2 + π 3 × π 6 − π 3 − 1 3 ( π 3 − π 6 ) = 2 + π 2 18 − 7 π 6 3 \begin{aligned} \xi \left(1\right) & = 2 + \frac \pi 2 \times 0 - \frac \pi 2 - 0 = 2 - \frac \pi 2 \\ \xi \left(\frac 12 \right) & = 2 + \frac \pi 6 \times \frac \pi 3 - \pi - \sqrt 3 \left(\frac \pi 6 - \frac \pi 3\right) = 2 + \frac {\pi^2}{18} - \pi + \frac \pi{2\sqrt 3} \\ \xi \left(\frac 1{\sqrt 2} \right) & = 2 + \frac \pi 4 \times \frac \pi 4 - \frac \pi{\sqrt 2} - \left(\frac \pi 4 - \frac \pi 4\right) = 2 + \frac {\pi^2}{16} - \frac \pi{\sqrt 2} \\ \xi \left(\frac {\sqrt 3}2 \right) & = 2 + \frac \pi 3 \times \frac \pi 6 - \frac \pi{\sqrt 3} - \frac 1{\sqrt 3} \left(\frac \pi 3 - \frac \pi 6\right) = 2 + \frac {\pi^2}{18} - \frac {7\pi}{6\sqrt 3} \end{aligned}

Therefore, ξ ( 1 ) + ξ ( 1 2 ) + ξ ( 1 2 ) + ξ ( 3 2 ) = 8 + 25 144 π 2 − 3 2 π − 4 2 + 6 3 6 6 π \xi \left(1\right) + \xi \left(\dfrac 12 \right) + \xi \left(\dfrac 1{\sqrt 2} \right) + \xi \left(\dfrac {\sqrt 3}2 \right) = 8 + \dfrac {25}{144}\pi^2 - \dfrac 32 \pi - \dfrac {4\sqrt 2+6\sqrt 3}{6\sqrt 6} \pi .

⟹ a + b + c + d + 3 e + f + g = 8 + 25 + 144 + 3 + 3 ( 2 ) + 4 + 6 = 196 = 14 \implies \sqrt{a+b+c+d+3e+f+g} = \sqrt{8+25+144+3+3(2)+4+6} = \sqrt{196} = \boxed{14} .

2 Helpful 1 Interesting 2 Brilliant 0 Confused

@Naren Bhandari , once again I don't know what is the purpose of making the finding the final answer be so difficult. If you want your problem to be popular, the finding of the final answer should be easy. The final step to get the answer is just to ensure the solution is correct and we want it to be in integer so that solver cannot resort to calculator to get the decimal answer. It is not for the purpose of testing the skill of the solvers not to make mistakes or testing their patience. It is better off the solver moves on to another problem to learn more. More than half of the time spent just to get the answer while the important part is for solver to learn about the calculus part of integration.

Chew-Seong Cheong - 2 years, 4 months ago
Felipe Lorenzzon
Feb 1, 2019

Essentially, I used the same method of Chew-Seong Cheong. The only difference is that, in the end, I didn't get that bigger fraction with a square root on the denominator because I had rationalized everything. Then I had to solve a diophantine equation in order to go back to the irrational denominator! I found it very cool to use integers in a calculus problem. However, I think it was way too much handwork. There was no necessity of plugging in 4 different values for theta, just 1 or 2 would be enough since the hardest part sits on the integration process. Once you have the antiderivative, there's no difficulty on plugging in the values and simplifying the monstrous expression and you may actually forget some small detail and get the entire answer wrong.

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