# Math Help - Integral....

1. ## Integral....

Find $\int \frac{dx}{1+4sin^2x}$

thanks!

2. You can make varous subs to tackle this one.

$\int\frac{1}{1+4sin^{2}(x)}dx$

Let $x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du$

After you make the subs and do the algebra, you should get:

$\int\frac{1}{5u^{2}+1}du$

Now, make the sub: $u=\frac{tan(w)}{\sqrt{5}}, \;\ du=\frac{\sqrt{5}}{5cos^{2}(w)}dw$

Make the subs and it turns into a very friendly integral:

$\frac{1}{\sqrt{5}}\int{dw}$

Make the resubs and you're done.

3. Originally Posted by galactus
You can make varous subs to tackle this one.

$\int\frac{1}{1+4sin^{2}(x)}dx$

Let $x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du$

After you make the subs and do the algebra, you should get:

$\int\frac{1}{5u^{2}+1}du$
How did u simplify the expression after u made the substitution of $x=tan^{-1}(u)$? and how did you "see" this substitution?

4. The simplification is just simple algebra. You shouldn't have any trouble since you're at this level of mathematics. As for 'seeing' the sub, it's a matter of doing them, I supoose.

5. Originally Posted by polymerase
Find $\int \frac{dx}{1+4sin^2x}$
$\int {\frac{1}
{{1 + 4\sin ^2 x}}\,dx} = \int {\frac{1}
{{5\sin ^2 x + \cos ^2 x}}\,dx} = \int {\frac{{\sec ^2 x}}
{{5\tan ^2 x + 1}}\,dx} .$

And this is an arctangent, so

$\int {\frac{1}
{{1 + 4\sin ^2 x}}\,dx} = \frac{1}
{{\sqrt 5 }}\arctan \left( {\sqrt 5 \tan x} \right) + k.$

6. Originally Posted by galactus
The simplification is just simple algebra. You shouldn't have any trouble since you're at this level of mathematics. As for 'seeing' the sub, it's a matter of doing them, I supoose.
Thanks, i got it...i just fail to see at first that $u=tan\:x$! LOL