- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

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

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

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

- Admin
- #2

- Jan 26, 2012

- 4,198

- Admin
- #3

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

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

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!

- Thread starter
- Admin
- #4

- Feb 14, 2012

- 3,909