Integrate sin^34x

To integrate sin^34x, also written as ∫sin³4x dx, sin cubed 4x, sin^3(4x), and (sin 4x)^3, we start by using standard trig identities to simplify the integral.

We factor out one of the sin4x terms, and therefore the integration remains the same, except that we now have a sin squared term.

We remember the Pythagorean trig identity and rearrange it for the sin squared term for the purposes of substituting the expression into our integration problem.

As you can see, after substitution, we now have a different expression, which means the same thing.

We let u = cos4x.

Then du/dx = -4sin4x. I show how to do this differentiation at the bottom of the page, if you need help with it.

We rearrange the expression for du.

As you can see, our integration problem has sin4x dx, which almost looks the same as the expression for du. The only issue we have is with -4, otherwise we would be able to make a straight swap. To solve this issue, we play a little mathematical trick as shown below.

If we were to multiply -¼ with -4 then the result would be +1 and we have not changed anything. However, inside the integral, we now have -4sin4x dx which we can swap out with du.

We also swap out cos4x with u to give the expression above.

On the LHS we multiply out the -¼, which also results in sign reversal. We then integrate each term separately as shown on the RHS.

Hence, this is the integration solution in terms of u. We now need to replace u with cos4x.

Hence, this is the final answer.

Differentiate cos4x