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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.

- Thread starter gigi9
- Start date

- 41

- 0

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.

- 637

- 0

#1: It is the ∫ Sec^{-1}x * 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

Use parts formula: let u = Sec

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:

Science Advisor

Homework Helper

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sin

sin(θ)= √(1- cos

sec

sec(θ)= √(1+ tan

tan

tan(θ)= √(sec

When you see something like ∫dx/(x

a√((x/a)

x/a= sec(θ) (0r x= a sec(θ)) will work!

Doing that, √(x

Also, x

The entire integral becomes ∫(a sec(θ)tan(θ)dθ/(a

= (1/a

= (1/a

which can be done by using the trig identity:

cos

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