00:01
For this problem, we need to find the indefinite integral, using the substitution x equals tangent theta for the integral of 9x cube over the square root of 1 plus x squared dx.
00:13
So here we let x equals tangent theta, and we get the differential of x, that'll be dx equal to secan squared theta d theta.
00:25
And so from here we have the integral of 9x cube over the square root of 1 plus x squared dx.
00:34
This is equal to the integral of 9 times the cube of tangent theta all over the square root of 1 plus the square of tangent theta times d x which is secan squared theta d theta.
00:54
So simplifying we get 9 times tangent cube theta.
01:00
Over the square root of 1 plus tangent squared theta times secan squared theta d theta.
01:09
Now 1 plus tangent squared is just secan squared theta and so we have integral of 9 tangent cubed theta over the square root of secan squared theta times secan squared theta d theta.
01:27
Now this cancels out and we're left with one secan theta here and so we have 9 times the integral of tangent cube theta times secantheta d theta...