1.7 Integrals Resulting in Inverse Trigonometric Functions 

[STARTS AT 2:53]

PROFESSOR: This looks like it is in the form of also inverse sine. So in this case, our a is 1 and our u is 4x squared. Now, back here, with this problem here, we did not have to worry about our chain rule because u and x, we just switched the variable. Now here, we are going to go ahead and switch these. du is going to be 8x dx. 

All right, I made a mistake there. Go back. This is actually 4x, which makes this only-- makes it 4. So this is 4 dx, which makes 1/4 du equal dx. So we're going to replace this-- I'm going to rewrite the integral. So it is going to be 1/4 du over the square root of-- in this case, remember, a is just 1. So it is 1 squared minus 4x squared. So the integral of this is 1/4 sine inverse of u over a, so 4x over 1, plus C.

[ENDS AT 4:38]