00:01
For this, probably what we want to verify is the antiderivative of the square of sine inverse of x with respect to x is x times the square of sine inverse of x minus 2x plus 2 squared of 1 minus x squared sine of x or sine inverse of x plus c.
00:19
So first, we want to apply substitution to our integral.
00:24
Let's say y is equal to sine inverse of x.
00:32
Then, sine of y equals x and from here, if we take the derivative of both sides, cosine of y dy equals dx.
00:47
So, the integral of sine inverse of x squared dx equals the integral of y squared times dx which is cosine of y dy.
01:06
And then from here, we apply integration by parts.
01:09
We want to set u equal to y squared and dv equal to cosine of y dy.
01:17
We get du equal to 2y dy and v is going to equal sine of y.
01:31
So then from here, we have u times v that's y squared sine of y minus the integral of v du.
01:39
That's 2y sine of y dy and then we will apply again integration by parts...