# How do you find the antiderivative of int sin^3xcosxdx?

Nov 23, 2016

$\text{ "intsin^3xcosxdx" } = \frac{1}{4} {\sin}^{4} x + C$

#### Explanation:

no need for substitution here if you recognise that

$y = {\sin}^{n} x \implies \frac{\mathrm{dy}}{\mathrm{dx}} = n {\sin}^{n - 1} x \cos x \text{ }$using the chain rule

so$\text{ "intsin^3xcosxdx" }$suggests a function of the type

$y = {\sin}^{4} x$

lets check this by differentiating.

$u = \sin x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \cos x$

$y = {u}^{4} \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {u}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {u}^{3} \cos x = 4 {\sin}^{3} x \cos x$

$\text{ "intsin^3xcosxdx" } = \frac{1}{4} {\sin}^{4} x + C$

Nov 23, 2016

${\sin}^{4} \frac{x}{4} + C$.

#### Explanation:

Since you have a cosine terms hanging around some sine terms, it might be helpful to try the substitution $u = \sin x$, $\mathrm{du} = \cos x \mathrm{dx}$.

Using this substitution, $\int {\sin}^{3} x \cos x \mathrm{dx} = \int {u}^{3} \mathrm{du}$.

$\int {u}^{3} \mathrm{du} = {u}^{4} / 4 + C = {\sin}^{4} \frac{x}{4} + C$.