When it comes to integrals with 2 trigonometric functions, always play around with the Pythagorean Identities first.
For int35sin^4xcos^3xdx, start by pulling the 35 out of the integral to get 35intsin^4xcos^3xdx. Now, note that cos^3x=cos^2xcosx:
35intsin^4xcos^2xcosxdx
Apply the Pythagorean Identity cos^2x=1-sin^2x:
35intsin^4x(1-sin^2x)cosxdx
Distribute the cosx:
35intsin^4x(cosx-cosxsin^2x)dx
And distribute the sin^4x:
35intsin^4xcosx-sin^6xcosxdx
Finally, apply the sum rule to integrals to end up with:
35intsin^4xcosxdx-35intsin^6xcosxdx.
You might be wondering how what we just did is even remotely helpful. Well, if you look hard enough, you'll see that both of these integrals can be solved using a u-substitution. We will do these one by one, starting with 35intsin^4xcosxdx:
Let u=sinx->(du)/dx=cosx->du=cosxdx
With this substitution, the integral becomes:
35intu^4du
color(white)(XX)=7u^5+C_1
Because u=sinx,
35intsin^4xcosxdx=7sin^5x+C_1
On to 35intsin^6xcosxdx:
Let u=sinx->(du)/dx=cosx->du=cosxdx
With this substitution, the integral becomes:
35intu^6du
color(white)(XX)=5u^7+C_2
Because u=sinx,
35intsin^6xcosxdx=5sin^7x+C_2
Our solution looks like:
int35sin^4xcos^3xdx=7sin^5x+C_1-(5sin^7x+C_2)
color(white)(XX)=7sin^5x+C_1-5sin^7x-C_2
You can combine C_1-C_2 into a general constant C:
int35sin^4xcos^3xdx=7sin^5x-5sin^7x+C
