Integral trig

**usagi_killer** · January 26th 2010, 03:15 AM

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

How do I evaluate this?

**Suhada** · January 26th 2010, 03:43 AM

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

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

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

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

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

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

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

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

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

**Archie Meade** · January 26th 2010, 03:59 AM

Originally Posted by usagi_killer

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

How do I evaluate this?

Let $u=x$

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

$w=Sinx,\ dw=Cosxdx$

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

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

Evaluating the resulting integral...

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

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

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

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

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