
#1
Nov803, 09:18 AM

P: 43

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^2a^2)) 2) indefinite integral (14x^2)^(3/2)dx How to use integration by part for indefinite integral of x*inverse tan(x) dx. 



#2
Nov803, 09:45 AM

P: 640

#1: It is the ∫ Sec^{1}x * x^{2} dx.
Use parts formula: let u = Sec^{1}, du = x'/xsqrt(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 



#3
Nov903, 07:00 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

The point of "trigonometric substitution" is to use trig identities such as
sin^{2}(θ)= 1 cos^{2}(θ) so that sin(θ)= √(1 cos^{2}(θ)) sec^{2}(θ)= 1+ tan^{2}(θ) so that sec(θ)= √(1+ tan^{2}(θ)) tan^{2}(θ)= sec^{2}(θ) 1 so that tan(θ)= √(sec^{2}(θ) 1) When you see something like ∫dx/(x^{3}√(x^{2}a^{2})) you should immediately think "Hmmm, that √(x^{2} a^{2}) reminds me of √(sec^{2}(θ) 1) especially if I factor out a^{2} 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, √(x^{2} a^{2}) becomes √(a^{2}sec^{2}(θ) a^{2})= a√(sec^{2}1)= a√(tan^{2})= a tan(θ)! Also, x^{3} become a^{3}sec^{3}(θ) and dx= a d(sec(θ))= a sec(θ)tan(θ) dθ The entire integral becomes ∫(a sec(θ)tan(θ)dθ/(a^{3}sec^{3}(θ)(a tan(θ))) = (1/a^{3})∫(1/sec^{2}(θ) = (1/a^{3})∫cos^{2}(θ)dθ which can be done by using the trig identity: cos^{2}(θ)= (1/2)(1+ cos(2θ)). 



#4
Nov903, 02:18 PM

P: 1,572

Trig subtitution
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 x^{3}, you'd have a^{3}a=a^{4} 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...



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