# What is the integral of sin(x)*sin(10x)?

Apr 22, 2015

Use one of the product to sum formulas. Not many of us use them enough to memorize them, but they are easy enough to get.

We want to change a product of sines. We recall the sine and cosine of a sum or difference.

For this problem we'll need

$\cos \left(a + b\right) = \cos a \cos b - \sin a \sin b$
$\cos \left(a - b\right) = \cos a \cos b + \sin a \sin b$

Subtracting the first equation fron the second gives us:

$\cos \left(a - b\right) - \cos \left(a + b\right) = 2 \sin a \sin b$

So $\sin a \sin b = \frac{1}{2} \cos \left(a - b\right) - \frac{1}{2} \cos \left(a + b\right)$

In this problem $a = x$ and $b = 10 x$

$\sin x \sin 10 x = \frac{1}{2} \cos \left(- 9 x\right) - \frac{1}{2} \cos 11 x$

Since cosine is an even function, make it
$\frac{1}{2} \cos 9 x - \frac{1}{2} \cos 11 x$

Now integrate each term by simple substitution.

Notes

To change a product of cosines, add the two formulas above.
$\cos a \cos b = \frac{1}{2} \cos \left(a - b\right) + \frac{1}{2} \cos \left(a + b\right)$

To change $\sin a \cos b$, write the formulas for $\sin \left(a + b\right)$ and $\sin \left(a - b\right)$ and add.