I need help on this integral

**Locke** · February 23rd 2010, 04:42 PM

Never mind. I got the solution.

**Prove It** · February 23rd 2010, 05:03 PM

Originally Posted by Locke

1 / (9 + x^2)^(3/2) dx

I actually got the answer to this integral here, but I have no idea how to get it (I need to know the steps).

Any ideas?

$\int{\frac{1}{(9 + x^2)^{\frac{3}{2}}}\,dx}$ .

Make the substitution $x = 3\sinh{t}$ so that $dx = 3\cosh{t}\,dt$ .

It's important to note that from our original substitution: $\sinh{t} = \frac{x}{3}$ .

So the integral becomes

$\int{\frac{1}{[9 + (3\sinh{t})^2]^{\frac{3}{2}}}\,3\cosh{t}\,dt}$

$= \int{\frac{3\cosh{t}}{[9(1 + \sinh^2{t})]^{\frac{3}{2}}}\,dt}$

$= \int{\frac{3\cosh{t}}{27(\sqrt{\cosh^2{t}})^3}\,dt }$

$= \frac{1}{9}\int{\frac{\cosh{t}}{\cosh^3{t}}\,dt}$

$= \frac{1}{9}\int{\frac{1}{\cosh^2{t}}\,dt}$

$= \frac{1}{9}\tanh{t} + C$

$= \frac{\sinh{t}}{9\cosh{t}} + C$

$= \frac{\sinh{t}}{9\sqrt{1 + \sinh^2{t}}} + C$ .

$= \frac{\frac{x}{3}}{9\sqrt{1 + \left(\frac{x}{3}\right)^2}} + C$

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

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

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

$= \frac{x}{9\sqrt{9 + x^2}} + C$ .

Another alternative is to make the substitution $x = 3\tan{\theta}$ .

**Locke** · February 23rd 2010, 05:10 PM

What does sinh mean? I did it by substituting x for 3tan and got the same answer.

**Prove It** · February 23rd 2010, 05:12 PM

Originally Posted by Locke

What does sinh mean? I did it by substituting x for 3tan and got the same answer.

The Hyperbolic sine.

Hyperbolic function - Wikipedia, the free encyclopedia

And like I said, substituting $x = 3\tan{\theta}$ should give you the same answer.

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