# Thread: Help Getting Started on: 9dx/Sqrt(81-x^4)

1. ## Help Getting Started on: 9dx/Sqrt(81-x^4)

Not sure how to get this one going exactly, we just started getting into Integrals that really require manipulation before solving them and I believe this is definitely one of those! Thanks in advance.

$\int{9dx/\sqrt{81-x^4}}$

2. This seems like a rather messy problem, are you sure you have written it correctly?

Anyway, you could rewrite the nominator as:

$\frac{9}{\sqrt{3-x} \sqrt{x+3} \sqrt{x^2+9}}$

But I akm not sure how much of a helpt this would be.

3. Yeah I just went ahead and double checked, the final answer is fairly complicated, it includes a tan function as well.

4. Right I'll post working as I go along... Haven't checked this yet so just making it up as I go...

Let $x = 3\sin(u)$.

then $dx = 3\cos(u)du$.

So out integral becomes...

$\int \frac{9 \cdot 3\cos(u)}{\sqrt{81 - 81\sin^4(u)}}du$

$= \int \frac{27\cos(u)}{\sqrt{81(1 - \sin^4(u))}}du$

$= \int \frac{27\cos(u)}{9\sqrt{cos^4(u)}}du$

$= \int \frac{27\cos(u)}{9\cos^2(u)}du$

5. $= \int \frac{3}{\cos(u)}du$

$= 3\int \frac{1}{\cos(u)}du$

$= 3\int \sec(u) du$.

$= 3\ln|\sec(u) + \tan(u)|$. (proof http://math2.org/math/integrals/more/sec.htm)

6. Thanks a bunch, could you explain the first part to me? When we do trig subs our professor always makes us use a triangle like noobs so I'm a little confused.

(Anyone else could chime in as well!)

Thanks again!!!