Question

Integration question, see inside.

Answer 1

\[\pi\int\limits\limits_{0}^{\pi} \sin ^{2}x dx\]

Answer 2

write it as: (1-cos2x)/2

Answer 3

or you can use integration by parts or the reduction formulae

but the re-writing it is easier

Answer 4

After I rewrite it, would using the substitution method be the easiest way to solve from there?

Answer 5

\[\int \frac{1-cos2x}{2}dx\]

Answer 6

\[\int \frac{1}{2}(1-cos2x)dx\]
you should be able to do it now

and dont forget about the limits!

Answer 7

so, integrate by parts: \[\int\limits_{}^{}(1/2) dx - \int\limits_{}^{} (\cos2x)/2 dx\]

Answer 8

oh no no

why would you want to use integration by parts now?

Answer 9

I am having trouble with the later integration.

Answer 10

do you want to use integration by parts for this integral or not?

Answer 11

\[\frac{1}{2} \int(1-cos2x)\]
can you see that this is a standard integral?

Answer 12

Not if I don't have to - Oh shoot! you can pull that half out infront, can't you

Answer 13

Yes, I totally over looked that, and tried multiplying the half in.