00:01
In this problem we are given some integrals and we are going to propose or find appropriate trigonometric substitutions to simplify these integrals but we are not going to evaluate these integrals just the substitution rules.
00:19
Okay so if we look at the questions each of the integrals we will see some common pattern here and i want to mention this before actually solving them.
00:31
So if our integral contains something like x squared minus a squared to some power, then the rule of time says the proper substitution is x equal to a times second of t or some other variable.
00:54
If we have something like x squared plus a squared to some power n, then the proper substitution is a.
01:04
Times tangent t finally if we have something like a squared minus x squared to some power and this time the substitution becomes a times sine t okay with these in mind now let us uh solve the so all these questions okay part a we have integral t x for x squared minus two to the power three halves.
01:43
This is in the first form or we can reduce this to the first one.
01:49
So it has factor out four here.
01:55
Four to the power three halves x squared minus two over four or one over two.
02:05
So this guy is a squared so we should have x equal to root one half times second of t...