Solve integral \intx^p/(x\sqrt(1-x^2))dx?
I recognize that \int1/\sqrt(1-x^2) involves trigonometric substitution, but I don't know how to take the other x 's out
I recognize that
1 Answer
int \ (x^p)/(xsqrt(1-x^2)) \ dx = int \ sin^(p-1) theta \ d theta
And:
int \ sin^nx \ dx = -1/nsin^(n-1)x \ cosx + (n-1)/n \ int \ sin^(n-2)x \ dx
Explanation:
We seek:
I = int \ (x^p)/(xsqrt(1-x^2)) \ dx
Consider:
I_n =int \ x^n/(sqrt(1-x^2)) \ dx
We can perform a substitution:
x = sin theta => dx/(d theta) = cos theta
And if we substitute into the integral, we get:
I_n = int \ (sin theta)^n/(sqrt(1-(sin theta)^2)) \ cos theta \ d theta
\ \ \ = int \ (sin^n theta)/(sqrt(1 - sin^2 theta)) \ cos theta \ d theta
\ \ \ = int \ (sin^n theta)/(sqrt(cos^2 theta)) \ cos theta \ d theta
\ \ \ = int \ sin^n theta \ d theta
Then we can write:
I = int \ (x^(p-1))/(sqrt(1-x^2)) \ dx
\ \ = I_(p-1)
\ \ = int \ sin^(p-1) theta \ d theta
And, a reduction formula can be used for the resultant integral.
int \ sin^nx \ dx = -1/nsin^(n-1)x \ cosx + (n-1)/n \ int \ sin^(n-2)x \ dx