# [SOLVED] Trig Substitution Integral

• Feb 12th 2009, 06:32 PM
mollymcf2009
[SOLVED] Trig Substitution Integral
Integrate.

$\int \frac{3}{x^2\sqrt{100-sin x^2}} dx$

Here is what I have done (I left out all my early steps):

$\int \frac{30 cos\theta d\theta}{(10sin\theta)^2 (\sqrt{100-sin ^2\theta)}}$

$
\int \frac{30 cos\theta d\theta}{100(sin^2\theta) \cdot (10\sqrt{1-sin ^2 \theta}}$

$
\int \frac{30 cos\theta d\theta}{1000sin^2\theta \sqrt{cos ^2 \theta}}$

$= \frac{30}{1000} \int \frac{1}{sin^2 \theta} d\theta$

$
= \frac{30}{1000} \int \frac{2}{1-cos(2\theta)} d\theta$

Can someone please check this for me? Should I do a u sub with $2\theta$? Thanks!!
• Feb 12th 2009, 07:17 PM
Jameson
I think you went one line too far. $\frac{1}{\sin^2(x)} = \csc^2(x)$, which is an easy integral.
• Feb 12th 2009, 07:17 PM
Chop Suey
Just to make sure, I will assume that you meant
$\int \frac{3}{x^2\sqrt{100-x^2}}~dx$

In your second step, you probably forgot to write $100\sin^2{\theta}$.

In your last step, you applied a trig identity in a wrong way. $\frac{1}{\sin^2{x}}$ is a standard form. If it helps, $\frac{1}{\sin^2{x}} = \csc^2{x}$