Integration By Parts

[Brain0991]

Find the Volume of the solid obtained by revolving the region bounded by y=sin(x), x=0 and x=pi/2

pi?sin[sup:2sdu2uuw]2[/sup:2sdu2uuw]x dx

I decided not to let u=sinx and dv=sinxdx because I would just get cos[sup:2sdu2uuw]2[/sup:2sdu2uuw]x in the next integral and I think I would be going in circles

u=sin[sup:2sdu2uuw]2[/sup:2sdu2uuw]x du=2cosxsinxdx
dv=dx v=x

uv-?vdu
xsin[sup:2sdu2uuw]2[/sup:2sdu2uuw]x - 2?xsinxcosx dx

and then I am stuck

 

[galactus]

When integrating \(\displaystyle sin^{2}(x)\) there is no need for parts.

Use the identity \(\displaystyle sin^{2}(x)=\frac{1}{2}-\frac{1}{2}cos(2x)\)

Now it is easier. No powers to deal with.

 

[Brain0991]

Unfortunately for me, my teacher asked this question on our homework and he is forcing us to do integration by parts for it, so was I on the right track with what I did before doing it with integration by parts?

p.s. I do not understand why my teacher would do that either.

 

[Deleted member 4993]

Unfortunately for me, my teacher asked this question on our homework and he is forcing us to do integration by parts for it, so was I on the right track with what I did before doing it with integration by parts? .....NO

p.s. I do not understand why my teacher would do that either.

 

[galactus]

I can't see the reasoning, but if you have to use parts note that a reduction formula can be derived for powers of sin(x).

\(\displaystyle \int sin^{n}(x)dx=\frac{-1}{n}sin^{n-1}(x)cos(x)+\frac{n-1}{n}\int sin^{n-2}(x)dx\)

In your case, n=2.

This can be done using parts by letting \(\displaystyle u=sin^{n-1}(x), \;\ dv=sin(x)dx, \;\ du=(n-1)sin^{n-2}(x)(cos(x))dx, \;\ v=\int sin(x)dx=-cos(x)\)