Integral with trig functions?

you want to write it as (sinx)^6(cosx)^5 = (sinx)^5((cosx)^2)^2*cosx = (sinx)^5(1-(sinx)^2)^2 and do a u-substitution where u=sinx. So du=cosxdx. Then you have the integral of u^5(1-u^2)^2du =

integral of u^5(1-2u^2+u^4)du =

integral of u^5-2u^7+u^9 du=

1/6 u^6 - 1/4 u^8 + 1/10 u^10 +c

Now plug sinx back in for u and you are done.

You always want to convert the trig function with an odd power and leave the other one alone, that way you end up with exactly one sin or cos and the rest are all sin or cos so the u-sub works. If they are both to odd powers, pick whichever one you want

If they are both to even powers, cry. Not really, but it is more work, you have to use the half-angle identities to convert them all to sin2x and cos2x. Then apply the same rules. Good luck!

∫ sin^6 x cos^5 x dx = ∫ sin^6 x cos^4 x cos x dx = ∫ sin^6 x (1 - sin^2 x)^2 cos x dx = ∫ sin^6 x (1 - 2 sin^2 x + sin^4 x) cos x dx

Let u = sin x and du = cos x dx

∫ (u^6 - 2 u^8 + u^10) du = u^7/7 - 2u^9/9 + u^11/11 = sin^7 x/7 - 2 sin^9 x/9 + sin^11 x/11 + C