Question

integral from sin^5xdx is?
please explain how you got your answer so i can undestand the probelm. thanks.. :D

Answer 1

\[\Large\bf\sf \int\limits \sin^5x\;dx\]We gotta solve this bad boy?

Answer 2

Oh it's an `odd` exponent, so this will work out nicely!

Answer 3

\[\Large\bf\sf \color{#DD4747 }{\sin^2x\quad=\quad 1-\cos^2x}\]We'll apply this identity to our problem.\[\large\bf\sf \int\limits\limits \sin^5x\;dx\quad=\quad \int\limits \sin x \sin^2x \sin^2x\;dx\quad=\quad \int\limits \sin x(\color{#DD4747 }{\sin^2x})^2\;dx\]

Answer 4

\[\Large\bf\sf \int\limits \sin x(\color{#DD4747 }{1-\cos^2x})^2\;dx\]

Answer 5

From here we can make a nice easy u-substitution.
\[\Large\bf\sf u=\cos x\]

Answer 6

Think you can solve it from there?
Shouldn't be too bad after you get your du.

Answer 7

\[\int\limits_{}^{}(1-u^{2})^{2}sinxdx\]

Answer 8

\[\int\limits_{}^{}(1-U ^{2})^{2}dU\]

Answer 9

isn't it?

Answer 10

thx

Answer 11

derivative of cosx is -sinx.
I think our substitution should bring along a negative sign in our du.

Answer 12

so..
\[-\int\limits_{}^{}(1-u ^{2})^{2}du\]
\[-\int\limits_{}^{}1-2u ^{2}+u ^{4}du\]

Answer 13

how about the sinx one?