# Integral with trig substitution

[SOLVED] Integral with trig substitution

1. Homework Statement

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

2. 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

3. 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!

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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.

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!

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

dextercioby
Homework Helper
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
Homework Helper
… just simplify …

Integration by parts not necessary:

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

Polynomial division?

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