# Thread: Integral trig

1. ## Integral trig

$\displaystyle \int x\sin^2(x)\cos(x)dx$

How do I evaluate this?

2. $\displaystyle I=\int x\sin^2(x)\cos(x)dx$

Use product rule: $\displaystyle \int uv' = uv -\int u'v$

$\displaystyle Let \ u=x \ \ so: \ \ u'=1$

$\displaystyle Let \ v'=sin^2xcosx \ \ so: \ \ v=\frac{1}{3}sin^3x$

$\displaystyle I=uv - \int u'v = \frac{1}{3}xsin^3x - J$
where
$\displaystyle J=\frac{1}{3}\int sin^3xdx$

$\displaystyle =\frac{1}{3}\int sinx(1-cos^2x)dx$

$\displaystyle =\frac{1}{3}\int sinx-cos^2xsinxdx$

$\displaystyle =\frac{1}{3}((-cosx)-(-\frac{1}{3}cos^3x)) + C$

$\displaystyle I=\frac{1}{3}(xsin^3x+cosx-\frac{1}{3}cos^3x) + C$

3. Originally Posted by usagi_killer
$\displaystyle \int x\sin^2(x)\cos(x)dx$

How do I evaluate this?
Let $\displaystyle u=x$

$\displaystyle dv=Sin^2xCosxdx=w^2dw$

$\displaystyle w=Sinx,\ dw=Cosxdx$

$\displaystyle v=\frac{w^3}{3}$

$\displaystyle \int{xSin^2xCosx}dx=x\ \frac{Sin^3x}{3}-\frac{1}{3}\int{w^3}dx=x\ \frac{Sin^3x}{3}-\frac{1}{3}\int{Sin^3x}dx$

Evaluating the resulting integral...

$\displaystyle \int{Sin^3x}dx=\int{Sin^2xSinx}dx=\int{(1-Cos^2x)Sinx}dx$

let $\displaystyle u=Cosx,\ -du=Sinxdx$

gives $\displaystyle -\int{(1-u^2)}du=\int{(u^2-1)}du=\frac{1}{3}u^2-u+C=\frac{1}{3}Cos^3x-Cosx+C$

$\displaystyle \int{xSin^2xCosx}dx=x\ \frac{Sin^3x}{3}-\frac{1}{9}Cos^3x+\frac{1}{3}Cosx+C$