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!