How do you find the integral of sin^3(x) cos^5(x) dx?
1 Answer
May 15, 2016
Explanation:
Recall that through the Pythagorean Identity
Thus,
intsin^3(x)cos^5(x)dx=intsin(x)(1-cos^2(x))cos^5(x)dx
Distributing just the cosines, this becomes
=int(cos^5(x)-cos^7(x))sin(x)dx
Now use the substitution:
Noting that
=-int(u^5-u^7)du
Integrating, this becomes
=-(u^6/6-u^8/8)+C
Reordering and back-substituting with
=cos^8(x)/8-cos^6(x)/6+C
Note that this integration could have also been done my modifying the cosines like:
cos^5(x)=cos(x)(cos^2(x))^2=cos(x)(1-sin^2(x))^2
And then proceeding by expanding and letting