00:01
Okay, this is a trigonometric substitution problem.
00:02
So x is equal to 7 sine of x, i mean sine of theta.
00:12
So dx will be 7 cosine of theta, d theta.
00:20
Okay? and also x squared will be 49 sine squared theta.
00:28
So let's plug everything in.
00:29
49 sine squared theta over 49, 49 minus 49 sine squared theta.
00:45
And d theta is 7 cosine theta, d theta.
00:50
So this is 49 cosine squared theta squared, so this is 7 cosine theta.
00:57
That goes away.
01:00
So we have 49 sine squared theta, d theta.
01:06
And sine squared theta is, by the double angle identities, 1 minus cosine 2 theta over 2.
01:15
Okay? so then we're going to have 49 over 2 minus 49 cosine 2 theta over 2, d theta...