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

Mar 8, 2015

The antiderivative of ${\cos}^{3} \left(x\right) \sin \left(x\right) \sqrt{1 - {\cos}^{4} \left(x\right)}$
is $\left(\frac{1}{5}\right) {\left(1 - {\cos}^{4} x\right)}^{\frac{5}{4}} + C$ which could also be written:

$\left(\frac{1}{5}\right) {\left(\sqrt{1 - {\cos}^{4} x}\right)}^{5} + C$

The method of solving may depend on your notation:

Method 1:
Let $u = 1 - {\cos}^{4} x$
so that $\mathrm{du} = - 4 {\cos}^{3} x \left(- \sin x\right) \mathrm{dx} = 4 {\cos}^{3} x \sin x \mathrm{dx}$.

Therefore, ${\cos}^{3} x \sin x \mathrm{dx} = \frac{\mathrm{du}}{4}$.

Using integral notation for the antiderivative, we can write:

$\int {\cos}^{3} \left(x\right) \sin \left(x\right) \left(\sqrt{1 - {\cos}^{4} \left(x\right)}\right) \mathrm{dx}$

$= \int {\left(1 - {x}^{4}\right)}^{\frac{1}{4}} {\cos}^{3} \left(x\right) \sin \left(x\right) \mathrm{dx} = \int {u}^{\frac{1}{4}} \frac{\mathrm{du}}{4}$

$= \frac{1}{4} \int {u}^{\frac{1}{4}} \mathrm{du} = \frac{1}{4} \frac{{u}^{\frac{5}{4}}}{\frac{5}{4}} + C = \left(\frac{1}{5}\right) {\left(1 - {\cos}^{4} x\right)}^{\frac{5}{4}} + C$

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

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

The function you asked about is almost ${\left(g \left(x\right)\right)}^{\frac{1}{4}} g ' \left(x\right)$.
It is exactly $\left(\frac{1}{4}\right) {\left(g \left(x\right)\right)}^{\frac{1}{4}} g ' \left(x\right)$.
So the antiderivative is $\left(\frac{1}{4}\right) {\left(g \left(x\right)\right)}^{\frac{5}{4}} / \left(\frac{5}{4}\right) + C$

which is the same as $\left(\frac{1}{5}\right) {\left(1 - {\cos}^{4} x\right)}^{\frac{5}{4}} + C$

As always, you should check the answer by differentiating.