Integration Grandmaster - P1

Calculus Level 5

Given that F ( x ) = ∫ cos ⁡ ( sin ⁡ x ) + cos ⁡ 2 x 1 + sin ⁡ x sin ⁡ ( sin ⁡ x ) d x F(x) = \displaystyle \int \dfrac{\cos(\sin x) + \cos^2 x}{1 + \sin x \sin(\sin x)} dx , F ( π ) = 0 F(\pi)=0 , find F ( 1 ) F(1) .

Submit ⌊ 10000 F ( 1 ) ⌋ \lfloor 10000 F(1) \rfloor .


The answer is -17939.

Alice Smith by Alice Smith , 20, China

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1 solution

Mark Hennings
Dec 10, 2020

Make the substitution y = cos ⁡ ( 1 2 ( x − sin ⁡ x ) ) sin ⁡ ( 1 2 ( x + sin ⁡ x ) ) y \; = \; \frac{\cos\big(\tfrac12(x - \sin x)\big)}{\sin\big(\tfrac12(x + \sin x)\big)} Then

1 + y 2 = sin ⁡ 2 ( 1 2 ( x + sin ⁡ x ) ) + cos ⁡ 2 ( 1 2 ( x − sin ⁡ x ) ) sin ⁡ 2 ( 1 2 ( x + sin ⁡ x ) ) = 1 − cos ⁡ ( x + sin ⁡ x ) + cos ⁡ ( x − sin ⁡ x ) + 1 2 sin ⁡ 2 ( 1 2 ( x + sin ⁡ x ) ) = 2 + cos ⁡ ( x − sin ⁡ x ) − cos ⁡ ( x + sin ⁡ x ) 2 sin ⁡ 2 ( 1 2 ( x + sin ⁡ x ) ) = 1 + sin ⁡ x sin ⁡ ( sin ⁡ x ) sin ⁡ 2 ( 1 2 ( x + sin ⁡ x ) ) \begin{aligned} 1 + y^2 & = \; \frac{\sin^2\big(\tfrac12(x + \sin x)\big) + \cos^2\big(\tfrac12(x - \sin x)\big)}{\sin^2\big(\tfrac12(x + \sin x)\big)} \; = \; \frac{1 - \cos(x + \sin x) + \cos(x - \sin x) + 1}{2\sin^2\big(\tfrac12(x + \sin x)\big)}\\ & = \; \frac{2 + \cos(x - \sin x) - \cos(x + \sin x)}{2\sin^2\big(\tfrac12(x + \sin x)\big)} \; = \; \frac{1 + \sin x \sin(\sin x)}{\sin^2\big(\tfrac12(x + \sin x)\big)} \end{aligned} while d y d x = − ( 1 − cos ⁡ x ) sin ⁡ ( 1 2 ( x − sin ⁡ x ) ) 2 sin ⁡ ( 1 2 ( x + sin ⁡ x ) ) − ( 1 + cos ⁡ x ) cos ⁡ ( 1 2 ( x − sin ⁡ x ) ) cos ⁡ ( 1 2 ( x + sin ⁡ x ) ) 2 sin ⁡ 2 ( 1 2 ( x + sin ⁡ x ) ) = − 1 2 sin ⁡ 2 ( 1 2 ( x + sin ⁡ x ) ) [ ( 1 − cos ⁡ x ) sin ⁡ ( 1 2 ( x − sin ⁡ x ) ) sin ⁡ ( 1 2 ( x + sin ⁡ x ) ) + ( 1 + cos ⁡ x ) cos ⁡ ( 1 2 ( x − sin ⁡ x ) ) cos ⁡ ( 1 2 ( x + sin ⁡ x ) ) ] = − cos ⁡ ( sin ⁡ x ) + cos ⁡ 2 x 2 sin ⁡ 2 ( 1 2 ( x + sin ⁡ x ) ) \begin{aligned} \frac{dy}{dx} & = \; -\frac{(1 - \cos x)\sin\big(\tfrac12(x-\sin x)\big)}{2\sin\big(\tfrac12(x + \sin x)\big)} - \frac{(1 + \cos x)\cos\big(\tfrac12(x - \sin x)\big)\cos\big(\tfrac12(x +\sin x)\big)}{2\sin^2\big(\tfrac12(x + \sin x)\big)} \\ & = \; -\frac{1}{2\sin^2\big(\tfrac12(x + \sin x)\big)}\left[ (1 - \cos x)\sin\big(\tfrac12(x - \sin x)\big)\sin\big(\tfrac12(x + \sin x)\big) + (1 + \cos x)\cos\big(\tfrac12(x - \sin x)\big)\cos\big(\tfrac12(x + \sin x)\big)\right] \\ & = \; -\frac{\cos(\sin x) + \cos^2x}{2\sin^2\big(\tfrac12(x + \sin x)\big)} \end{aligned} and hence F ( x ) = − 2 ∫ d y 1 + y 2 F(x) \; =\; -2\int \frac{dy}{1 + y^2} Since F ( π ) = 0 F(\pi) = 0 we deduce that F ( x ) = − 2 tan ⁡ − 1 ( cos ⁡ ( 1 2 ( x − sin ⁡ x ) ) sin ⁡ ( 1 2 ( x + sin ⁡ x ) ) ) F(x) \; =\; -2\tan^{-1}\left(\frac{\cos\big(\tfrac12(x - \sin x)\big)}{\sin\big(\tfrac12(x + \sin x)\big)} \right) which makes ⌊ 1 0 4 F ( 1 ) ⌋ = − 17939 \lfloor 10^4 F(1) \rfloor = \boxed{-17939} .

1 Helpful 0 Interesting 4 Brilliant 0 Confused

What is the motivation for this substitution?

Elijah L - 6 months ago

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... it works! I guess things like cos ⁡ ( x ± sin ⁡ x ) \cos(x\pm\sin x) and sin ⁡ ( x ± sin ⁡ x ) \sin(x\pm\sin x) are ways of getting terms like cos ⁡ ( sin ⁡ x ) \cos(\sin x) and sin ⁡ x sin ⁡ ( sin ⁡ x ) \sin x \sin(\sin x)

Mark Hennings - 6 months ago

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