integration by parts

**qzno** · March 22nd 2009, 04:39 PM

use integration by parts to find:

$\int^{\frac{\pi}{2}}_{0} xsin^2xcosx dx$

**skeeter** · March 22nd 2009, 05:05 PM

Originally Posted by qzno

use integration by parts to find:

$\int^{\frac{\pi}{2}}_{0} xsin^2xcosx dx$

$u = x$

$du = dx$

$dv = \sin^2{x}\cos{x} \, dx$

$v = \frac{\sin^3{x}}{3}$

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

for the last integral ...

$\sin^3{x} = (1 - \cos^2{x})\sin{x} = \sin{x} + \cos^2{x}(-\sin{x})$

**qzno** · March 22nd 2009, 05:07 PM

$dv = \sin^2{x}\cos{x} \, dx$

$v = \frac{\sin^3{x}}{3}$

is this a rule?

**skeeter** · March 22nd 2009, 05:22 PM

Originally Posted by qzno

$dv = \sin^2{x}\cos{x} \, dx$

$v = \frac{\sin^3{x}}{3}$

is this a rule?

it's an antiderivative using informal substitution ... note that $\cos{x}$ is the derivative of $\sin{x}$ ...

$\sin^2{x}\cos{x} \, dx$ is the same as $t^2 dt$

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