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Evaluate Definite Integral Challenge

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anemone

MHB POTW Director
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Feb 14, 2012
3,677
Evaluate \(\displaystyle \int_0^{\pi} \frac{\cos 4x-\cos 4 \alpha}{\cos x-\cos \alpha} dx\)
 

Ackbach

Indicium Physicus
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Jan 26, 2012
4,191
I'm sure everyone here would consider it cheating, but my handy-dandy HP 50g yields
$$ \int_{0}^{ \pi} \frac{ \cos(4x)- \cos(4 \alpha)}{ \cos(x)- \cos( \alpha)} \, dx=8 \pi \cos^{3}( \alpha)-4 \pi \cos( \alpha)=4 \pi \cos( \alpha)[2 \cos^{2}( \alpha)-1]=4 \pi \cos( \alpha) \cos(2 \alpha).$$
 

MarkFL

Administrator
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Feb 24, 2012
13,775
My solution:

If we apply a double-angle identity for cosine on the numerator of the integrand, we find:

\(\displaystyle \cos(4x)-\cos(4\alpha)=\left(2\cos^2(2x)-1 \right)-\left(2\cos^2(\alpha)-1 \right)=\)

\(\displaystyle 2\left(\cos(2x)+\cos(2\alpha) \right)\left(\cos(2x)-\cos(2\alpha) \right)\)

Applying a double-angle identity for cosine again, we have:

\(\displaystyle 4\left(\cos(2x)+\cos(2\alpha) \right)\left(\cos(x)+\cos(\alpha) \right)\left(\cos(x)-\cos(\alpha) \right)\)

Hence, the definite integral may now be written:

\(\displaystyle I=4\int_0^{\pi}\left(\cos(2x)+\cos(2\alpha) \right)\left(\cos(x)+\cos(\alpha) \right)\,dx\)

Using the property \(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\) we also have:

\(\displaystyle I=4\int_0^{\pi}\left(\cos(2x)+\cos(2\alpha) \right)\left(-\cos(x)+\cos(\alpha) \right)\,dx\)

Adding the two expressions for $I$, we obtain:

\(\displaystyle 2I=8\cos(\alpha)\int_0^{\pi}\cos(2x)+\cos(2\alpha)\,dx\)

Hence:

\(\displaystyle I=4\pi\cos(\alpha)\cos(2\alpha)\)

Adrian, glad to see we obtained the same result! :D
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,677
Thanks for participating, Ackbach and MarkFL and I think both of you deserve a pat on the back for this prompt reply...though Ackbach did depend on a little gizmo to obtain the answer, hehehe...