# Thread: stuck on integration

1. ## stuck on integration

How do you integrate:
sin(x)sin(nx)dx between Pi and Pi/2 when n doesnt equal 1

by parts doesnt work does it?
any help would be great

2. Originally Posted by ben.mahoney@tesco.net
How do you integrate:
sin(x)sin(nx)dx between Pi and Pi/2 when n doesnt equal 1

by parts doesnt work does it?
any help would be great
Remember that:

$\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ (1)

$\cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ (2)

Now (2)-(1) gives:

$\cos(A-B) - \cos(A+B) = 2\sin(A)\sin(B)$

Hence:

$\sin(A)\sin(B) = \frac{ \cos(A-B) - \cos(A+B)}{2}$

Which means

$\sin(x)\sin(nx) = \frac{ \cos(x - nx) - \cos(x+nx)}{2} = \frac{ \cos(x(1-n)) - \cos(x(1+n))}{2}$

So your integral becomes

$\int_{\frac{\pi}{2}}^{\pi} \frac{ \cos(x(1-n)) - \cos(x(1+n))}{2}\,dx = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \cos(x(1-n))\,dx - \frac{1}{2}\int_{\frac{\pi}{2}}^{\pi} \cos(x(1+n)) \,dx$

Which should be fairly simple if you remember that $\int \cos(kx) \,dx = \frac{\sin(kx)}{k} + C$

3. Originally Posted by ben.mahoney@tesco.net
How do you integrate:
sin(x)sin(nx)dx between Pi and Pi/2 when n doesnt equal 1

by parts doesnt work does it?
any help would be great
Use the fact that

$\sin nx \sin mx = \frac{1}{2} \cos (m-n) x - \frac{1}{2} \cos (m+n) x$

4. One possibility is the application of the famous formula...

$\sin \alpha \cdot \sin \beta = \frac{1}{2} \cdot \{ \cos (\alpha - \beta) - \cos (\alpha + \beta)\}$

Kind regards

$\chi$ $\sigma$

5. thanks everyone
got it now