Find $\displaystyle \int \frac{dx}{1+4sin^2x}$
thanks!
You can make varous subs to tackle this one.
$\displaystyle \int\frac{1}{1+4sin^{2}(x)}dx$
Let $\displaystyle x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du$
After you make the subs and do the algebra, you should get:
$\displaystyle \int\frac{1}{5u^{2}+1}du$
Now, make the sub: $\displaystyle u=\frac{tan(w)}{\sqrt{5}}, \;\ du=\frac{\sqrt{5}}{5cos^{2}(w)}dw$
Make the subs and it turns into a very friendly integral:
$\displaystyle \frac{1}{\sqrt{5}}\int{dw}$
Make the resubs and you're done.
$\displaystyle \int {\frac{1}
{{1 + 4\sin ^2 x}}\,dx} = \int {\frac{1}
{{5\sin ^2 x + \cos ^2 x}}\,dx} = \int {\frac{{\sec ^2 x}}
{{5\tan ^2 x + 1}}\,dx} .$
And this is an arctangent, so
$\displaystyle \int {\frac{1}
{{1 + 4\sin ^2 x}}\,dx} = \frac{1}
{{\sqrt 5 }}\arctan \left( {\sqrt 5 \tan x} \right) + k.$