Question

How to integrate this

Answer 1

To integrate, let's first clean out the inside, and simplify:

$16\int {(sin x)^4} \ dx$

We can now safely substitute the value of u:

$u = sin x$

$du = cos (x) dx$

Now, we can write that:

$16\int{u^4} \ du$

This is much easier to integrate:

$16\int{u^4} \ du = 16[\frac{1}{5}u^5] + C$

Simplify:

$\frac{16}{5}u^5 + C$

Replace u (final answer):

$\frac{16}{5}sin^5x +C$

Hope I could help with your integration.