Integral with trig substitution

[paralian]

[SOLVED] Integral with trig substitution

Homework Statement

Find \int(x^3)/\sqrt{x^2-9}

Homework Equations

Trig substitution. sin^2 + cos^2 =1, and other things that you can figure out from that.

Half angle formula, cos^2\theta=(1+cos(2\theta) )*.5

The Attempt at a Solution

Let x=3*sec\theta
so dx=3*sec\theta*tan\theta d\theta

When I substitute that in and simplify it, I got:

27*\int(sec^4(\theta) d\theta)

And I don't know how to integrate that. Half angle formulas aren't seeming to work.

Thanks!

 

[Tedjn]

Or, you can transform some of the secants into tangents:

27\int \sec^4 \theta d\theta = 27\int (1+\tan^2 \theta)\sec^2 \theta d\theta

and you can go from there.

 

[paralian]

Or, you can transform some of the secants into tangents:

27\int \sec^4 \theta d\theta = 27\int (1+\tan^2 \theta)\sec^2 \theta d\theta

and you can go from there.

27\int (1+\tan^2 \theta)\sec^2 \theta d\theta

27\int \sec^2 \theta d\theta + 27\int \tan^2 \theta\sec^2 \theta d\theta

27*tan\theta + 9\tan^3 \theta

Haha sorry I kind of forgot that \int \sec^2\theta d\theta was tan!

 

[Tedjn]

You are almost done. Because the problem was originally in terms of x, you will need to transform \theta back into x.

 

[dextercioby]

There's no need for trigonometry. Part integration and an immediate substitution will do the trick.

\int \frac{x^{3}}{\sqrt{x^{2}-9}}dx=\int x^{2}\frac{x}{\sqrt{x^{2}-9}}dx=\allowbreak x^{2}\sqrt{x^{2}-9}-2\int x\sqrt{x^{2}-9}\,dx=\allowbreak x^{2}\sqrt{x^{2}-9}-\frac{2}{3}\left( \sqrt{x^{2}-9}\right) ^{3}

up to a constant of integration.

 

[tiny-tim]

… just simplify …

Integration by parts not necessary:

\int \frac{x^{3}}{\sqrt{x^{2}-9}}dx<br /> =\int x\sqrt{x^{2}-9}dx\,+\,\int \frac{9x}{\sqrt{x^{2}-9}}dx<br /> \,=\,\frac{1}{3}(x^{2}-9)^{3/2}\,+\,9\sqrt{x^{2}-9}<br /> \,=\,(\frac{1}{3}x^2\,+\,6)\sqrt{x^{2}-9}\,.

 

[rocomath]

Polynomial division?

 

[paralian]

You are almost done. Because the problem was originally in terms of x, you will need to transform \theta back into x.

Right...I always forget that!

x=3*sec \theta \Rightarrow \theta = Sec^-1 (x/3)

\Rightarrow tan\theta = \sqrt{x^2-9}/3

9 \sqrt{x^2-9} + (x^2-9)^\frac{3}{2} /3

Shiny! Thanks