Evaluate Definite Integral Challenge

[anemone]

Evaluate \(\displaystyle \int_0^{\pi} \frac{\cos 4x-\cos 4 \alpha}{\cos x-\cos \alpha} dx\)

 

[Ackbach]

I'm sure everyone here would consider it cheating, but my handy-dandy HP 50g yields

Spoiler

$$ \int_{0}^{ \pi} \frac{ \cos(4x)- \cos(4 \alpha)}{ \cos(x)- \cos( \alpha)} \, dx=8 \pi \cos^{3}( \alpha)-4 \pi \cos( \alpha)=4 \pi \cos( \alpha)[2 \cos^{2}( \alpha)-1]=4 \pi \cos( \alpha) \cos(2 \alpha).$$

 

[MarkFL]

My solution:

Spoiler

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

 

[anemone]

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...

 

[CellCoree]

To evaluate the definite integral, we can use the trigonometric identity:
cos(a) - cos(b) = -2sin((a+b)/2)sin((a-b)/2)
Applying this identity to the integrand, we get:
\frac{\cos 4x-\cos 4 \alpha}{\cos x-\cos \alpha} = \frac{-2\sin(4x+4\alpha)/2 \sin(4x-4\alpha)/2}{-2\sin(x+\alpha)/2 \sin(x-\alpha)/2}
= \frac{\sin(2x+2\alpha) \sin(2x-2\alpha)}{\sin(x+\alpha) \sin(x-\alpha)}

We can then use the double angle formula for sine:
\sin(2x) = 2\sin(x)\cos(x)
to simplify the integrand further:
\frac{\sin(2x+2\alpha) \sin(2x-2\alpha)}{\sin(x+\alpha) \sin(x-\alpha)} = \frac{2\sin(x)\cos(x+2\alpha) \cdot 2\sin(x)\cos(x-2\alpha)}{\sin(x+\alpha) \sin(x-\alpha)}
= \frac{4\sin^2(x)\cos(x+2\alpha)\cos(x-2\alpha)}{\sin(x+\alpha)\sin(x-\alpha)}

Next, we can use the sum-to-product formula for cosine:
\cos(a)\cos(b) = \frac{1}{2}(\cos(a+b) + \cos(a-b))
to simplify the integrand even further:
\frac{4\sin^2(x)\cos(x+2\alpha)\cos(x-2\alpha)}{\sin(x+\alpha)\sin(x-\alpha)} = \frac{4\sin^2(x)}{2}\left(\frac{\cos(3x+2\alpha)+\cos(x+2\alpha)}{\sin(x+\alpha)} + \frac{\cos(3x-2\alpha)+\cos(x-2\alpha)}{\sin(x-\alpha)}\right)

We can now split the integral into two parts and use the substitution u = x + \alpha for the first integral and u = x - \alpha for the second integral:
\int_0^{\pi} \frac{\