How to use trig substitution for integrals with square roots

[gigi9]

Can someone please 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.

 

[PrudensOptimus]

#1: It is the ∫ 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

 

[HallsofIvy]

The point of "trigonometric substitution" is to use trig identities such as
sin²(θ)= 1- cos²(θ) so that
sin(θ)= √(1- cos²(θ))

sec²(θ)= 1+ tan²(θ) so that
sec(θ)= √(1+ tan²(θ))

tan²(θ)= sec²(θ)- 1 so that
tan(θ)= √(sec²(θ)- 1)

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

Doing that, √(x²- a²) becomes √(a²sec²(θ)- a²)= a√(sec²-1)= a√(tan²)= a tan(θ)!

Also, x³ become a³sec³(θ) and dx= a d(sec(θ))= a sec(θ)tan(θ) dθ

The entire integral becomes ∫(a sec(θ)tan(θ)dθ/(a³sec³(θ)(a tan(θ)))
= (1/a³)∫(1/sec²(θ)
= (1/a³)∫cos²(θ)dθ
which can be done by using the trig identity:
cos²(θ)= (1/2)(1+ cos(2θ)).

 

[phoenixthoth]

i was thinking about that and actually shouldn't it be |atanθ|, not atanθ? the a vs |a| part doesn't matter because with the x³, you'd have a³|a|=a⁴ anyway but there'd still be a |tanθ|. perhaps when the integration is done and you go back to x, it won't matter if it was tanθ or |tanθ|...