Did I solve this trig integral correctly?

[Guest]

Here's the problem: S sin^3(x)cos(x)dx [S means Integral]

Here's my solution:

S sin^2(x)sin(x)cos(x)dx
S (1-cos^2(x))sin(x)cos(x)dx
let u=cox(x), du=-sin(x)dx
-S (1-u^2)u du
-S u-u^3 du
-[(1/2)u^2-(1/4)u^4]+C

[(-1/2)(cos^2(x))+(1/4)(cos^4(x))]+C

Thanks!

 

[pka]

Don't make it hard!
That is just \(\displaystyle \L \int {u^3 du}.\)

 

[skeeter]

you sure took the long road to Dallas ...

\(\displaystyle \L \int \sin^3{x} \cos{x} dx\)

let u = sinx, du = cosx dx, substitute ...

\(\displaystyle \L \int u^3 du = \frac{u^4}{4} + C = \frac{\sin^4{x}}{4} + C\)

 

[Guest]

WOW... Well, it was in the section about separating trig integrals, i.e. if sine is odd and positive, save one sine and convert the rest to cosines... so I guess that's what I was trying to do? Thanks for the corrections!