# Math Help - inverse trig substitution?

1. ## inverse trig substitution?

Integrate:
x^3 dx/(sqrt (9+x^2))

I did this by drawing triangles and then using substitution but I ended up with the wrong answer. Help anyone? Thanks.

2. Hello, solars!

$\int\frac{x^3}{\sqrt{9+x^2}}\,dx$

Let: $x \:=\:3\tan\theta \quad\Rightarrow\quad dx \:=\:3\sec^2\!\theta\,d\theta$

Substitute: . $\int\frac{(27\tan^3\!\theta)(3\sec^2\!\theta\,d\th eta)}{3\sec\theta} \;=\;27\int\sec\theta\tan^3\!\theta\,d\theta$

. . $= \;27\int\tan^2\!\theta(\sec\theta \tan\theta\,d\theta) \;= \;27\int(\sec^2\!\theta-1)(\sec\theta\tan\theta \,d\theta)$

Let: $u \:=\:\sec\theta \quad\Rightarrow\quad du \:=\:\sec\theta\tan\theta\,d\theta$

Substitute: . $27\int(u^2-1)\,du \;=\;27\left(\frac{u^3}{3} - u\right) + C \;=\;9u(u^2-3) + C$

Back-substitute: . $9\sec\theta(\sec^2\!\theta - 3) + C$

Back-substitute again.
We had: . $x \:=\:3\tan\theta \quad\Rightarrow\quad \tan\theta \:=\:\frac{x}{3} \:=\:\frac{opp}{adj}$

$\theta$ is in a right triangle with: . $opp = x,\; adj = 3$
Using Pythagorus, we get: . $hyp = \sqrt{9+x^2}$
Hence: . $\sec\theta \:=\:\frac{hyp}{adj} \:=\:\frac{\sqrt{9+x^2}}{3}$

Substitute: . $9\cdot\frac{\sqrt{9+x^2}}{3}\left(\frac{9+x^2}{9} - 3\right) + C \;=\;3\sqrt{9+x^2}\left(\frac{9+x^2-27}{9}\right) + C$

Answer: . $\frac{1}{3}\!\cdot\!\sqrt{9+x^2}\!\cdot\!(x^2-18) + C$

3. Just put $z^2=9+x^2.$

4. Put z^2 as 9+x^2 then differentiate both sides to obtain xdx=zdz.

now substitute the values in the problem to the integrand as {z^2-9}/z use simple division and integrate.Resubstitute the value of x.

and most important trick of all. never use trigonometric substitution for such problem. At least not in the exams.You will find yourself in a terrible mess

5. As per usual I arrive late to a forum and all the good methods are already taken.

To add my own touch of "creative flair", this is what I did:

It might be easier solving the expression in two sections. It also becomes easier to do using interation by parts, but the substitution u^2=x^2+9 will work rather quickly.

Hope this helps!