00:01
Since this problem wants us to evaluate the definite integral using two ways, the first one would be to use the given integration limits, and the other one would be to use the limits obtained by trigonometric substitution, then it would be wise to find the general antiderivative first.
00:24
So in here we have, in general, the integral of t squared over, minus t squared raised to 3 over 2 d t.
00:37
Now in here we can apply trigonometric substitution because this follows the format of a squared minus x squared which uses x equal to a sine theta.
00:52
So in here we let t equal to sine theta so that d t is cosine theta d t is cosine theta d theta.
01:04
So from here, we have the integral of t squared that'll be sine squared theta over we have one minus sine squared theta raised two three over two times d t which is just cosine theta d theta now from here we simplify and we get the integral of sine squared theta times cosine theta over we have one minus sine squared which is equal to cosine squared theta.
01:39
So cosine squared theta raised to 3 over 2, that would be cosine cube theta and then d theta.
01:47
Now we can cancel out a cosine theta and we are left with a cosine squared theta and the denominator.
01:55
And so we have the integral of sine squared theta over cosine squared theta, which is the same as tangent squared theta d theta.
02:04
Now tangent squared is the same as secund squared theta minus 1 d theta and so integrating we have tangent theta minus theta plus c.
02:19
Now theta is not the original variable.
02:23
So for the first part we need to get the form of this in terms of t.
02:31
So we go back to the substitution.
02:35
If t is sine theta, then we say theta must be sine inverse of t.
02:47
And we can draw a right triangle with theta here.
02:52
Now sine theta is the same as opposite over hypotenus.
02:56
So the opposite side is t.
02:58
Hypotenus is one...