# How can I integrate sin^2 xcosx??

Mar 29, 2018

The integral is equal to $\frac{1}{3} {\sin}^{3} x + C$

#### Explanation:

Let $u = \sin x$. Then $\mathrm{du} = \cos x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{\cos} x$.

$I = \int {u}^{2} \cos x \cdot \frac{\mathrm{du}}{\cos} x$

$I = \int {u}^{2} \mathrm{du}$

$I = \frac{1}{3} {u}^{3} + C$

$I = \frac{1}{3} {\sin}^{3} x + C$

Hopefully this helps!

Mar 29, 2018

$\frac{1}{3} {\sin}^{3} x + C$

#### Explanation:

Given: $\int {\sin}^{2} x \cos x \mathrm{dx}$

Use $u$-substitution.
Let u = sin x; " "du = cos x dx; " "dx = (du)/(cos x)

$\int {\sin}^{2} x \cos x \mathrm{dx} = \int {u}^{2} \cancel{\cos x} \frac{\mathrm{du}}{\cancel{\cos x}} = \int {u}^{2} \mathrm{du} = \frac{1}{3} {u}^{3} + C$

$\int {\sin}^{2} x \cos x \mathrm{dx} = \frac{1}{3} {\sin}^{3} x + C$

Mar 29, 2018

Is an inmediate integral. See below

#### Explanation:

$\int {\sin}^{2} x \cos x \mathrm{dx} = \frac{1}{3} {\sin}^{3} x + C$ because the derivative of ${\sin}^{3} x$ is $3 {\sin}^{2} x \cos x$. To remove $3$ we need to insert $\frac{1}{3}$ before

Mar 29, 2018

$\frac{1}{3} {\sin}^{3} x + c$

#### Explanation:

$\int {\sin}^{2} x \cos x \mathrm{dx}$

Make an appropriate u sub:

$u = \sin x$

$\mathrm{du} = \cos x \mathrm{dx}$

$\implies \int {u}^{2} \mathrm{du}$

$\implies \frac{1}{3} {u}^{3} + c$

$\implies \frac{1}{3} {\sin}^{3} x + c$