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

1 Answer
Jun 2, 2018

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