Question

What is the integral of `int sin^4(x) dx`?

Answer 1

`int\ sin^4(x)\ dx=3/8x-1/4sin(2x)+1/32sin(4x)+C`

Explanation:

This integral is mostly about clever rewriting of your functions. As a rule of thumb, if the power is even, we use the double angle formula. The double angle formula says:
`sin^2(theta)=1/2(1-cos(2theta))`

If we split up our integral like this,
`int\ sin^2(x)*sin^2(x)\ dx`

We can use the double angle formula twice:
`int\ 1/2(1-cos(2x))*1/2(1-cos(2x))\ dx`

Both parts are the same, so we can just put it as a square:
`int\ (1/2(1-cos(2x)))^2\ dx`

Expanding, we get:
`int\ 1/4(1-2cos(2x)+cos^2(2x))\ dx`

We can then use the other double angle formula
`cos^2(theta)=1/2(1+cos(2theta))`
to rewrite the last term as follows:
`1/4int\ 1-2cos(2x)+1/2(1+cos(4x))\ dx=`

`=1/4(int\ 1\ dx-int\ 2cos(2x)\ dx+1/2int\ 1+cos(4x)\ dx)=`

`=1/4(x-int\ 2cos(2x)\ dx+1/2(x+int\ cos(4x)\ dx))`

I will call the left integral in the parenthesis Integral 1, and the right on Integral 2.

Integral 1
`int\ 2cos(2x)\ dx`

Looking at the integral, we have the derivative of the inside, `2` outside of the function, and this should immediately ring a bell that you should use u-substitution.

If we let `u=2x`, the derivative becomes `2`, so we divide through by `2` to integrate with respect to `u`:
`int\ (cancel(2)cos(u))/cancel(2)\ du`

`int\ cos(u)\ du=sin(u)=sin(2x)`

Integral 2
`int\ cos(4x)\ dx`

It's not as obvious here, but we can also use u-substitution here. We can let `u=4x`, and the derivative will be `4`:
`1/4int\ cos(u)\ dx=1/4sin(u)=1/4sin(4x)`

Completing the original integral
Now that we know Integral 1 and Integral 2, we can plug them back into our original expression to get the final answer:
`1/4(x-sin(2x)+1/2(x+1/4sin(4x)))+C=`

`=1/4(x-sin(2x)+1/2x+1/8sin(4x))+C=`

`=1/4x-1/4sin(2x)+1/8x+1/32sin(4x)+C=`

`=3/8x-1/4sin(2x)+1/32sin(4x)+C`