Question

integrate sin^5(x) dx. i just need a little point in the right direction!

Answer 1

\[\sin ^5x~dx=\left( \sin ^2~x \right)^2 \sin xdx= \left( 1-\cos ^2x \right)^2\sin x~dx\]
pu t cos x=t

Answer 2

I assume you mean this:
\[\int\limits_{?}^{?}(\sin x)^5 dx\]

Is that correct?

Answer 3

oh yes, that is correct. i am not sure how to type that into the open questions, only when replying.

Answer 4

@surjithayer that is how i started it actually but i must be doing something wrong after this step

Answer 5

\[\int\limits \left( 1-\cos ^2x \right)^2\left( \sin x \right)dx\]
\[=\int\limits \left( 1-2\cos ^2x+\cos ^4x \right)\sin x~dx=-\cos x+\frac{ 2\cos ^3x }{ 3 }-\frac{ \cos ^5x }{ 5 }+c\]

Answer 6

or you can solve after substitution cosx=t
-sinx dx=dt
sin xdx=-dt

Answer 7

\[\left( 1-t^2 \right)^2=1-2t^2+t4\]

Answer 8

one brute force method is as follows
sin(x)=(ie^(-ix)-ie^(ix))/2
expand (ie^(-ix)-ie^(ix))^5 and integrate

Answer 9

i ended up learning from you guys and solved it properly! thanks!

Answer 10

Great! Thank you for acknowledging our help.

Answer 11

you are welcome.