# Evaluate Definite Integral Challenge

#### anemone

##### MHB POTW Director
Staff member
Evaluate $$\displaystyle \int_0^{\pi} \frac{\cos 4x-\cos 4 \alpha}{\cos x-\cos \alpha} dx$$

#### Ackbach

##### Indicium Physicus
Staff member
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

Staff member
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! #### anemone

##### MHB POTW Director
Staff member
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...