Question

What is the antiderivative of `cos^3(x) sin(x) (root(4){1-cos^4(x)})`?

Answer 1

The antiderivative of `cos^3(x) sin(x) root(4){1-cos^4(x)}`
is `(1/5)(1-cos^4x)^(5/4)+C` which could also be written:

`(1/5)(root(4)(1-cos^4x))^5+C`

The method of solving may depend on your notation:

Method 1:
Let `u=1-cos^4x`
so that `du=-4cos^3x(-sinx)dx=4cos^3xsinxdx`.

Therefore, `cos^3xsinxdx=(du)/4`.

Using integral notation for the antiderivative, we can write:

`intcos^3(x) sin(x) (root(4){1-cos^4(x)})dx`

`=int(1-x^4)^(1/4)cos^3(x) sin(x)dx=intu^(1/4)(du)/4`

`=1/4intu^(1/4)du=1/4(u^(5/4))/(5/4)+C=(1/5)(1-cos^4x)^(5/4)+C`

Method 2:
If you do not recognize the above notation, try this:

Let `g(x)=1-x^4`. If we took the derivative of `g(x)` to a power, we would need to multiply by `g'(x)` which is `4cos^3xsinx`.
(That is what the chain rule tells us.)

The function you asked about is almost `(g(x))^(1/4)g'(x)`.
It is exactly `(1/4)(g(x))^(1/4)g'(x)`.
So the antiderivative is `(1/4) (g(x))^(5/4)/(5/4)+C`

which is the same as `(1/5)(1-cos^4x)^(5/4)+C`

As always, you should check the answer by differentiating.