00:01
In this question, we are asked to calculate the given definite integral.
00:04
And to do that, we are going to use the substitution x equals to 1 over 4 second theta.
00:15
Then d x equals to 1 over 4 second theta times tangent theta, d theta.
00:31
And then we can rewrite the integral as the integral from square root 2 over 4 to 2 over 4.
00:38
Note that i haven't changed the limits of integration here, which means that when we get the answer, we'll have to get back to the old variables.
00:50
So, dx changes into one quarter, second theta times tangent theta, and x equals to one quarter second theta.
01:08
And under the square root, we are going to get 16 times 1 over 16, second squared theta minus 1.
01:23
And d theta.
01:27
Now 16 cancels and basically that's the reason we did the substitution 1 quarter second theta.
01:33
Because now under the square root we've gotten second squared theta minus 1, which equals to tangent squared theta.
01:49
And we can rewrite the integral as also note that 1 quarter cancels and second theta also cancels.
02:02
And we're going to get the integral from square of 2 over 4 to 2 over 4 of tangent theta d theta over the square root of tangent squared theta which is going to be just tangent theta so tangent theta also cancels and we are going to get the interval from square 2 over 4 to 2 over 4 to 2 over 4 which is basically theta theta in substitution from square 2 over 4 to 2 over 4 now we need to get back to the old variables x equals to 1 quarter second theta...