# How do you evaluate \int \cos 3x \sin ^ { 2} x d x?

Mar 5, 2018

$\int \cos 3 x {\sin}^{2} x \mathrm{dx} = {\sin}^{3} \frac{x}{3} - \frac{4 {\sin}^{5} x}{5} + C$

#### Explanation:

Use the trigonometric formulas:

$\cos 3 x = \cos \left(2 x + x\right) = \cos 2 x \cos x - \sin 2 x \sin x$

$\cos 3 x = \left({\cos}^{2} x - {\sin}^{2} x\right) \cos x - 2 \cos x {\sin}^{2} x$

$\cos 3 x = {\cos}^{3} x - 3 \cos x {\sin}^{2} x$

so:

$\cos 3 x {\sin}^{2} x = {\cos}^{3} x {\sin}^{2} x - 3 \cos x {\sin}^{4} x$

$\cos 3 x {\sin}^{2} x = {\cos}^{3} x \left(1 - {\cos}^{2} x\right) - 3 \cos x {\sin}^{4} x$

$\cos 3 x {\sin}^{2} x = {\cos}^{3} x - {\cos}^{5} x - 3 \cos x {\sin}^{4} x$

Solve the integral separately:

$\int {\cos}^{3} x \mathrm{dx} = \int \cos x \left(1 - {\sin}^{2} x\right) \mathrm{dx}$

$\int {\cos}^{3} x \mathrm{dx} = \int \cos x \mathrm{dx} - \int {\sin}^{2} x \cos x \mathrm{dx}$

$\int {\cos}^{3} x \mathrm{dx} = \sin x - \int {\sin}^{2} x d \left(\sin x\right)$

$\int {\cos}^{3} x \mathrm{dx} = \sin x - {\sin}^{3} \frac{x}{3} + {c}_{1}$

then:

$\int {\cos}^{5} x = \int \cos x {\left(1 - {\sin}^{2} x\right)}^{2} \mathrm{dx}$

$\int {\cos}^{5} x = \int \left(1 - 2 {\sin}^{2} x + {\sin}^{4} x\right) d \left(\sin x\right)$

$\int {\cos}^{5} x = \sin x - \frac{2 {\sin}^{3} x}{3} + {\sin}^{5} \frac{x}{5} + {c}_{2}$

and finally:

$\int \cos x {\sin}^{4} x \mathrm{dx} = \int {\sin}^{4} x d \left(\sin x\right) = {\sin}^{5} \frac{x}{5} + {c}_{3}$

Putting the partial solutions together:

$\int \cos 3 x {\sin}^{2} x \mathrm{dx} = \sin x - {\sin}^{3} \frac{x}{3} - \sin x + \frac{2 {\sin}^{3} x}{3} - {\sin}^{5} \frac{x}{5} - \frac{3 {\sin}^{5} x}{5} + C$

and simplifying:

$\int \cos 3 x {\sin}^{2} x \mathrm{dx} = {\sin}^{3} \frac{x}{3} - \frac{4 {\sin}^{5} x}{5} + C$

Mar 5, 2018

$I = {\sin}^{3} \frac{x}{3} - \frac{4 {\sin}^{5} x}{5} + C$

#### Explanation:

$\textcolor{red}{\int {\left[f \left(x\right)\right]}^{n} \frac{d}{\mathrm{dx}} f \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + C}$

$I = \int \left[4 {\cos}^{3} x - 3 \cos x\right] {\sin}^{2} x \mathrm{dx}$
$= \int \left[4 {\cos}^{3} x \cdot {\sin}^{2} x - 3 \cos x \cdot {\sin}^{2} x\right] \mathrm{dx}$$= \int \left[4 \cos x {\cos}^{2} x {\sin}^{2} x - 3 {\sin}^{2} x \cos x\right] \mathrm{dx}$$= \int \left[4 \cos x \left(1 - {\sin}^{2} x\right) {\sin}^{2} x - 3 {\sin}^{2} x \cos x\right] \mathrm{dx}$$= \int \left[4 {\sin}^{2} x \cos x - 4 {\sin}^{4} x \cos x - 3 {\sin}^{2} x \cos x\right] \mathrm{dx}$$= \int \left[{\sin}^{2} x \cos x - 4 {\sin}^{4} x \cos x\right] \mathrm{dx}$$= \int {\left(\sin x\right)}^{2} \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right) \mathrm{dx} - 4 \int {\left(\sin x\right)}^{4} \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right) \mathrm{dx}$$= {\left(\sin x\right)}^{3} / 3 - 4 \cdot {\left(\sin x\right)}^{5} / 5 + C$
$= {\sin}^{3} \frac{x}{3} - \frac{4 {\sin}^{5} x}{5} + C$