00:01
For problem 42, we're given an integral that we need to solve, but we can't solve it in this form.
00:06
So we're going to need to use u substitution and try to clean it up some.
00:09
So let's use u equals sine of the square root of theta.
00:14
And how did i come up with this? well, the derivative of sign is going to be cosine, right? and we'll also have to use the chain rule, which should produce something similar to this when it comes out front.
00:25
So by taking the derivative of this, we should be able to replace the entire rest of this integral as well as the original sign function.
00:35
So let's take that derivative and see how it works out.
00:37
D -u is going to become cosine of the square root of theta, d -theta, but chain rule, right? we need to multiply it by the derivative.
00:47
So the derivative of theta to the one -half power is going to be one -half theta to the negative one -half.
00:54
So let's go ahead and plug that in.
00:55
And i did not leave myself much space here.
01:01
One -half theta negative one -half, right? this is what du equals.
01:06
And you might be like, well, that doesn't look anything like that, but actually it does...