00:01
Evaluate the following integrals, a, the integral of x over the square root of 1 minus x to the 4th, dx.
00:09
So my goal here is to get rid of this variable on top.
00:13
By setting u equal to x squared, i can do exactly that by obtaining the derivative of 2x dx, or 1⁄2dx, or 1⁄2dx.
00:27
So we now have our substitutions.
00:32
Let's rewrite this integral as 1⁄2.
00:36
Over the square root 1 minus u squared d u.
00:42
Now we're going to use trigonometric substitutions to solve this from here.
00:48
I'm going to let u equal to sign of a new variable, we'll call it w.
00:54
The derivative of u will be cosine of w.
00:59
Now i can rewrite this integral as 1 half the integral of 1 over square root of 1 minus sine squared w times cosine of w...
