MHB Evaluate Definite Integral Challenge

AI Thread Summary
The discussion revolves around evaluating the definite integral $$\int_0^{\pi} \frac{\cos 4x-\cos 4 \alpha}{\cos x-\cos \alpha} dx$$. Participants express their thoughts on using calculators for solving the problem, with one member humorously acknowledging reliance on a calculator for the solution. There is a sense of camaraderie among contributors, with appreciation for prompt responses. The conversation highlights the balance between manual calculations and technological assistance in solving mathematical challenges. Overall, the thread emphasizes the collaborative nature of tackling complex integrals.
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Evaluate $$\int_0^{\pi} \frac{\cos 4x-\cos 4 \alpha}{\cos x-\cos \alpha} dx$$
 
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I'm sure everyone here would consider it cheating, but my handy-dandy HP 50g yields
$$ \int_{0}^{ \pi} \frac{ \cos(4x)- \cos(4 \alpha)}{ \cos(x)- \cos( \alpha)} \, dx=8 \pi \cos^{3}( \alpha)-4 \pi \cos( \alpha)=4 \pi \cos( \alpha)[2 \cos^{2}( \alpha)-1]=4 \pi \cos( \alpha) \cos(2 \alpha).$$
 
My solution:

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

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

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

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

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

Hence, the definite integral may now be written:

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

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

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

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

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

Hence:

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

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