But I See Two Variables!

Calculus Level 3

∫ 0 π / 6 cos ⁡ 2 x − cos ⁡ 2 β cos ⁡ x − cos ⁡ β d β = A π B cos ⁡ x + A \large \int_0^{\pi/6}\dfrac{\cos 2x-\cos 2\beta}{\cos x - \cos\beta}\, d\beta=\dfrac{A\pi}B\cos x+A

Let x x be a constant real number such that cos ⁡ x − cos ⁡ β ≠ 0 \cos x - \cos\beta \ne 0 for 0 ≤ β ≤ π 6 0 \leq \beta \leq \dfrac\pi 6 .

If the above equation is true for integers A A and B B , then find 2 ( A + B ) 2(A+B) .


The answer is 8.

Sravanth C. by Sravanth C. , 21, India

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1 solution

Sravanth C.
May 27, 2016

The most important thing is that there is only one variable β \beta , because the integral is with respect to β \beta and not x x , thus the expressions cos ⁡ 2 x \cos 2x and cos ⁡ x \cos x are just constant. Hence we can simplify the integral as follows;

∫ 0 π / 6 cos ⁡ 2 x − cos ⁡ 2 β cos ⁡ x − cos ⁡ β d β = ∫ 0 π / 6 2 cos ⁡ 2 x − 1 − ( 2 cos ⁡ 2 β − 1 ) cos ⁡ x − cos ⁡ β d β = ∫ 0 π / 6 2 ( cos ⁡ x + cos ⁡ β ) ( cos ⁡ x − cos ⁡ β ) cos ⁡ x − cos ⁡ β d β = ∫ 0 π / 6 ( 2 cos ⁡ x + 2 cos ⁡ β ) d β = ∫ 0 π / 6 2 cos ⁡ x d β + ∫ 0 π / 6 2 cos ⁡ β d β = 2 cos ⁡ x ⋅ β ∣ 0 π / 6 + 2 sin ⁡ β ∣ 0 π / 6 = 2 cos ⁡ x ⋅ π 3 + 2 ⋅ 1 2 = 1 3 π cos ⁡ x + 1 \begin{aligned} \int_0^{\pi/6}\dfrac{\cos 2x-\cos 2\beta}{\cos x - \cos\beta}\, d\beta &=\int_0^{\pi/6}\dfrac{2\cos ^2x-1-(2\cos ^2\beta-1)}{\cos x - \cos\beta}\, d\beta\\ &=\int_0^{\pi/6}\dfrac{2(\cos x+\cos\beta)(\cos x-\cos \beta)}{\cos x - \cos\beta}\, d\beta\\ &=\int_0^{\pi/6}(2\cos x+2\cos\beta) \, d\beta\\ &=\int_0^{\pi/6}\color{#D61F06}{2\cos x}\, \color{#3D99F6}{d\beta} \color{#333333}+\int_0^{\pi/6}\color{#D61F06}2\color{#3D99F6}{\cos\beta \, d\beta}\\ &=\color{#D61F06}{2\cos x}\cdot\color{#3D99F6}{\beta |_0^{\pi/6}}+\color{#D61F06}2\color{#3D99F6}{\sin\beta |_0^{\pi/6}}\\ &=\color{#D61F06}{2\cos x}\cdot\color{#3D99F6}{\dfrac\pi 3}+ \color{#D61F06}2\cdot\color{#3D99F6}{\dfrac 12}\\ &=\dfrac{1}3\pi\cos x + 1 \end{aligned}

Thus A = 2 A=2 and B = 3 B=3 , therefore 2 ( A + B ) = 2 ( 1 + 3 ) = 8 2(A+B)=2(1+3)=\boxed 8 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You made a couple of minor mistakes:

For your second line, it should be ∫ 0 π / 6 2 ( cos ⁡ x + cos ⁡ β ) ( cos ⁡ x − cos ⁡ β ) cos ⁡ x − cos ⁡ β d β \displaystyle \int_0^{\pi/6}\dfrac{2(\cos x+\cos\beta)(\cos x-\cos \beta)}{\cos x - \cos\beta}\, d\beta .

And here's where you made a couple of arithmetic mistakes.

⋯ = 2 cos ⁡ x ⋅ β ∣ 0 π / 6 + 2 sin ⁡ β ∣ 0 π / 6 = 2 cos ⁡ x ( π 6 − 0 ) + 2 ( sin ⁡ π 6 − sin ⁡ 0 ) = π 3 cos ⁡ x + 1 . \cdots = \color{#D61F06}{2\cos x}\cdot\color{#3D99F6}{\beta |_0^{\pi/6}}+\color{#D61F06}2\color{#3D99F6}{\sin\beta |_0^{\pi/6}} = 2 \cos x \left( \dfrac\pi 6 - 0\right) + 2 \left( \sin \dfrac\pi6 - \sin 0 \right) = \dfrac\pi3 \cos x + 1 \; .

Pi Han Goh - 5 years ago

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Stupid me. Thanks for pointing my blunders. I have edited the question accordingly, please check it now.

Sravanth C. - 5 years ago

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You don't need to write "coprime" in the problem statement. Other than that, you're right.

Don't worry about your blunders. We all make mistakes =D

Pi Han Goh - 5 years ago

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@Pi Han Goh – Done. Thanks again, hope you liked solving it.

Sravanth C. - 5 years ago

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