How do you evaluate the integral #int cos^3xsqrtsinx#?
1 Answer
Jan 21, 2017
Explanation:
Let
Then
#=>intcos^3xsqrt(u) * (du)/cosx#
#=>int cos^2x sqrt(u)du#
We can rewrite
#=>int (1 - sin^2x)sqrt(u)du#
Since
#=>int(1 - u^2)sqrt(u)du#
#=> int(1 - u^2)u^(1/2)du#
#=>int u^(1/2) - u^(5/2)du#
This can be evaluate as
#=>2/3u^(3/2) - 2/7u^(7/2) + C#
#=>2/3(sinx)^(3/2) - 2/7(sinx)^(7/2) + C#
Hopefully this helps!