00:02
In this problem, we are going to find integral of square root 1 minus 4x squared d of x using trigonometric substitution.
00:22
Now to get rid from the square root, we use the trigonometric substitution.
00:30
X is equal to 1 over 2 sine of theta.
00:35
This implies that dx will be equal to 1 over 2.
00:42
Differential of sine theta will be cosine of theta, d of theta.
00:49
Now, using these substitutions in the above integral, we have the given integral will be 1 minus 4 times.
01:01
Since x is 1 over 2 sine theta, so x squared will be 1 over 4, square theta whole square root d of x is 1 over 2 cost theta d theta now if we can sell 4 and 4 so we have 1 minus sine square theta which will be equal to cost square theta whole square root times 1 over 2 cos of theta d of theta now taking one over two outside the integral sign we have integral of here square and square root get cancelled and we have cost theta times cost theta is cost square theta d of theta now by trigonometric identities we know that cost square theta is one plus cos 2 theta over 2 d of theta.
02:25
Or this can also be written as 1 over 4 times integral of 1d theta plus 1 over 4 times integral of cost 2 theta d of theta...