00:01
For part c to evaluate this integral, the first thing we have to do is to apply trigonometric substitution.
00:05
In here, we want to set 2x equal to 3 secant theta, so that 4x squared is equal to 9 secant squared theta, and that dx is equal to 3 half secant theta, tangent theta, d theta.
00:33
This is equal to the integral of 1 over 9 secant squared theta minus 9.
00:42
This raised the 3 halves times 3 over 2 secant theta, tangent theta d theta.
00:52
And then we simplify, we have 3 halves times the integral of secant theta tangent theta all over.
01:03
We have 9 times secan squared theta minus 1 this whole thing raised to 3 over 2 and then d theta and since secan squared theta is tangent squared theta this is equal to 3 over 2 integral of secant theta theta tangent theta over 9 raised to 3 halves that's equal to 27 this times tangent theta raised through the power of 3, and then you have d -theta.
01:37
Can cancel out.
01:38
Tangent theta here, we have tangent squared theta left, and we simplify.
01:44
This is equal to 1 over 18 in a row of 1 over cosine theta times cosine squared theta over sine squared theta d -theta.
01:59
We cancel out cosine, if cosine here, and then we apply substitution.
02:06
We set u equal to sine theta.
02:11
D u is equal to cosine theta, d theta.
02:14
So this is 1 over 18 interval of d u over u squared.
02:22
This is equal to 1 over 18 times negative 1 over u.
02:27
Or that's negative 1 over 18 times sine theta plus c, or that's the same as negative 1 over 18 co -secent theta plus c...