Question

What parts do I need to integrate sin^(3x)*cos^2(x)dx? (Do I need to use algebraic substitution? If so what is the u value?)

Answer 1

Did you mean to type \(\Large\rm \sin^3x\cos^2x\) ?

Answer 2

yes

Answer 3

If you use your pythagorean identity, you can clean this up a little bit.
\[\Large\rm \sin x \sin^2x \cos^2x=\sin x(1-\cos^2x) \cos^2x\]
u= cos x
\[\Large\rm =-\int\limits (1-\cos^2x)\cos^2x (-\sin x~dx)\]

Answer 4

\[\Large\rm =-\int\limits\limits (1-u^2)u^2 (du)\]Understand what I did there?
Or was the thing with the negatives confusing?

Answer 5

Nope, the negatives were fine! I was trying the wrong u-value. Thank you so much!!

Answer 6

cool c:

Answer 7

If you were wondering about a method using integration by parts, you could split up the powers like zep did:
\[\int\sin^3x\cos^2x~dx=\int\sin^3 x\cos x\cos x~dx\]
Then set the following:
\[\begin{matrix}
u=\cos x&&&dv=\sin^3x\cos x~dx\\
du=-\sin x~dx&&&dv=t^3~dt&&\text{(where }t=\sin x)\\
&&&v=\frac{1}{4}\sin ^4 x
\end{matrix}\]
which gives
\[\int\sin^3x\cos^2x~dx=\frac{1}{4}\sin^4 x\cos x+\frac{1}{4}\int\sin^5x~dx\]
The new integral can be handled with another power split and some identity usage:
\[\int\sin x\sin^4x~dx=\int\sin x(1-\cos^2x)^2~dx=\int\sin x(1-2\cos^2x+\cos^4x)~dx\]
and so on. This way involves a lot of power reduction, many more steps than zep's clearly superior method. Always check for substitutions before considering IBP.