A calculus problem by Jose Sacramento

Calculus Level 4

∫ 0 1 ∫ 0 1 − x cos ⁡ ( x − y x + y ) d y d x \int_0^1 \int_0^{1-x} \cos \left( \dfrac{x-y}{x+y} \right) \, dy \; dx

If the integral above can be expressed as sin ⁡ A B \dfrac {\sin A}B , where A A and B B are both positive integers, find A + B A+B .


The answer is 3.

Jose Sacramento by Jose Sacramento , 66, Portugal

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1 solution

Apply u,v rule for double integral with x+y=u and x-y=v and multiply integral with the Jacobian J=|d(x,y)/d(u,v)| which turns out to be one eventually and hence the ans

1 Helpful 0 Interesting 0 Brilliant 0 Confused

u varies from 0 to 1 and v varies from -u to +u here

Chirag Shyamsundar - 4 years, 4 months ago

Can u show the limits of u and v ?

Kushal Bose - 4 years, 4 months ago

The problem here is that ∂ ( u , v ) ∂ ( x , y ) = ∣ 1 1 1 − 1 ∣ = − 2 \frac{\partial(u,v)}{\partial(x,y)} \; = \; \left| \begin{array}{cc} 1 & 1 \\ 1 & - 1 \end{array} \right| \; = \; -2 so that J = 2 J = 2 , not 1 1 .

The value of the integral is 1 2 sin ⁡ 1 \tfrac12\sin 1 . I have already reported this. If we introduce X = x + y X = x+y , the range of integration is 0 ≤ y ≤ X ≤ 1 0 \le y \le X \le 1 , and the integral is ∫ 0 1 d X ∫ 0 X d y cos ⁡ ( X − 2 y X ) = ∫ 0 1 [ − 1 2 X sin ⁡ ( X − 2 y X ) ] y = 0 X d X = sin ⁡ 1 ∫ 0 1 X d X = 1 2 sin ⁡ 1 \begin{aligned} \int_0^1\,dX \int_0^X\,dy\, \cos\left(\frac{X-2y}{X}\right) & = \; \int_0^1 \Big[-\tfrac12X \sin\left(\tfrac{X - 2y}{X}\right)\Big]_{y=0}^X\,dX \\ & = \; \sin 1 \int_0^1X\,dX \; = \; \tfrac12\sin 1 \end{aligned}

Mark Hennings - 4 years, 4 months ago

0 pending reports

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