How do you evaluate the integral #int cos^3xsin^3x#?
2 Answers
Explanation:
We want to rewrite so that either sine or cosine is all by itself with a power of
#=int cosx(cos^2xsin^3x)#
Rewrite
#=int cosx(1 - sin^2x)sin^3xdx#
#=int cosx(sin^3x - sin^5x)dx#
The derivative of
#=int cosx(u^3 - u^5) * (du)/cosx#
This will eliminate, leaving us only with u's.
#=int u^3 - u^5 du#
Integrate using
#=1/4u^4 - 1/6u^6 + C#
Resubstitute:
#=1/4(sinx)^4 - 1/6(sinx)^6 + C#
#=1/4sin^4x - 1/6sin^6x + C#
Hopefully this helps!
The answer is
Explanation:
Therefore,
So,