Question

using integration by substitution.....what is the integral of cosx sin^3 x dx??

Answer 1

what is u??3?

Answer 2

I'm thinking:
\[\int\limits_{}^{}\cos(x) \sin ^3(x)dx \rightarrow \int\limits_{}^{}\sin(x)\cos(x)\sin^2(x)dx\]\[\int\limits_{}^{}\sin(x)\cos(x)(1-\cos^2x)dx \rightarrow \int\limits_{}^{}\sin(x)(\cos(x)-\cos^3(x))dx\] So let u=cos(x),du=sin(x)dx
\[\int\limits_{}^{}(u-u^3)du \rightarrow 1/2u^2 -1/4u^4+c \rightarrow (1/2)\cos^2(x) - (1/4)\cos^4(x) + c\]

Answer 3

Naturally it was more simple than my route, by doing the path you were on you let u=sin(x), du=cos(x)dx and get\[\int\limits \cos(x)\sin^3(x)dx \rightarrow \int\limits u^3du \rightarrow \frac{u^4}{4}+c \rightarrow \frac{\sin^4(x)}{4}+c\]

Answer 4

okay so this is the correct answer??

Answer 5

Yes, seems you understand it better than you think. :)

Answer 6

guess so