Question

evaluate the integral of rsinx with respect to r and then to x, for the limits from r=0 to r=cos x and x=0 to x= pi...

Answer 1

\(\large \int \limits_0^{\pi}\int \limits _0^{\cos x}r \sin x drdx=\large \int \limits_0^{\pi} \sin x [\int \limits _0^{\cos x}r dr]dx\)
can you first evaluate inner integral ?

Answer 2

yes..after i evaluate the inner integral i get \[\frac{ 1 }{ 2 } \int\limits_{0}^{\Pi}sinxcos ^{2}xdx\]
so i should evaluate this w/ cos ^{2}x?

Answer 3

Yes. This time, you're integrating with x, and so, cos² x should be included, it is not a to be treated as a constant the way sin x was, when you were integrating with r :)

Answer 4

you put u= cos x, du = -sin x dx
integrating w.r.t x now..

Answer 5

thanks..the answer is 1/3..

Answer 6

yes.