Question

How do I do this integral? `int sin^3x cos^2x dx`

I tried to do it with half-angle formulas and got `1/4cosx - 1/16sin4x`, but it wasn't right... The book gives an answer of `cos^5x/5 - cos^3x/3 + C`.

Answer 1

As a general rule, try to find the simplest way to do a problem.

---

First, let's see how one could do the problem.

`color(blue)(int sin^3xcos^2xdx)`

`= intsinxsin^2cos^2xdx`

`= int sinx(1 - cos^2x)cos^2xdx`

`= int sinx(cos^2x - cos^4x)dx`

Let `u = cosx`, and then `du = -sinxdx`. Thus:

`=> -int -sinx(cos^2x - cos^4x)dx`

`= -int u^2 - u^4du`

`= -u^3/3 + u^5/5`

`= color(blue)(cos^5x/5 - cos^3x/3 + C)`

So I get the same answer the book gives when I do it that way.

I don't know what work you did, but the derivative of your answer doesn't return the original integral+-+1%2F16sin(4x)).

Also , you don't need to use the half-angle formula here (which are `sin(x/2) = pm sqrt((1-cosx)/2)` or `cos(x/2) = pm sqrt((1+cosx)/2)`); you don't have `trig(x/2)` anywhere.

And just so you know:

`cos^2x = (1+cos2x)/2`

`sin^2x = (1-cos2x)/2`

And if `u = cosx`, it's not a straightforward substitution to try to substitute for `cos2x`, as `cos2x = cosxcosx - sinxsinx`.

Also, since everything else was `trig(x)`, using those identities just makes it harder because now you strayed away from using `1*x` and now you have to get back using double angle identities.