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#
