# Trig subtitution

Can someone plz show me the concept of trig substituion and easy way to use it...I don't quite understand this...
I'm ot sure how to dothis problem below..
1) indefinite integral dx/(x^3sqrt(x^2-a^2))
2) indefinite integral (1-4x^2)^(3/2)dx

How to use integration by part for
indefinite integral of x*inverse tan(x) dx.

#1: It is the &int; Sec-1x * x-2 dx.

Use parts formula: let u = Sec-1, du = x'/|x|sqrt(x^2 - a^2) where a = 1. dv = x-2 dx, v = -x^-1.

The answer? I'm not sure, but out of scratch i got -sec^-1x/x + 2sqrt(x^6 - x^4)/6x^5 - 4x^3 + C = -Cosx/x - sinx * x^3/3

Last edited:
HallsofIvy
Homework Helper
The point of "trigonometric substitution" is to use trig identities such as
sin2(&theta;)= 1- cos2(&theta;) so that

sec2(&theta;)= 1+ tan2(&theta;) so that

tan2(&theta;)= sec2(&theta;)- 1 so that

When you see something like &int;dx/(x3&radic;(x2-a2)) you should immediately think "Hmmm, that &radic;(x2- a2) reminds me of &radic;(sec2(&theta;)- 1)- especially if I factor out a2 to get
a&radic;((x/a)2-1). I'll bet making a substitution like
x/a= sec(&theta;) (0r x= a sec(&theta;)) will work!