How do you integrate #int x^2*sin(2x) dx# from #[0,pi/2]#?

Question

How do you integrate #int x^2*sin(2x) dx# from  #[0,pi/2]#?

Paisley Johnson · Accepted Answer

The answer is #=pi^2/8-1/2#

First, calculate the indefinite integral by applying the integration by parts #2# times

#intuv'=uv-intu'v#

#u=x^2#, #=>#, #u'=2x#

#v'=sin2x#, #=>#, #v=-1/2cos2x#

The integral is

#I=intx^2sin2xdx=-x^2/2cos2x+intxcos2xdx#

#u=x#, #=>#, #u'=1#

#v'=cos2x#, #=>#, #v=1/2sin2x#

#intxcos2xdx=x/2sin2x-1/2intsin2xdx#

#=x/2sin2x+1/4cos2x#

Finally,

#I=-x^2/2cos2x+x/2sin2x+1/4cos2x#

And the definite integral is

#intx^2sin2xdx=[-x^2/2cos2x+x/2sin2x+1/4cos2x]_0^(pi/2)#

#=(pi^2/8-1/4)-(1/4)#

#=pi^2/8-1/2#

#=0.73#

Charles Arocha · Answer

undefined To integrate $ \int_{0}^{\frac{\pi}{2}} x^2 \sin(2x) \, dx $ from $0$ to $\frac{\pi}{2}$, you can use integration by parts. Let $ u = x^2 $ and $ dv = \sin(2x) \, dx $. Then, $ du = 2x \, dx $ and $ v = -\frac{1}{2} \cos(2x) $. Applying integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

You will obtain:

$$ \int_{0}^{\frac{\pi}{2}} x^2 \sin(2x) \, dx = \left[ -\frac{x^2}{2} \cos(2x) \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} -x \cos(2x) \, dx $$

Evaluate the limits and integrate the remaining term:

$$ \left[ -\frac{x^2}{2} \cos(2x) \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} -x \cos(2x) \, dx = \frac{\pi^2}{8} $$

So, $ \int_{0}^{\frac{\pi}{2}} x^2 \sin(2x) \, dx = \frac{\pi^2}{8} $.