1. ## integration

integarte {secx}\{1+2
sin^2(x)},dx

2. ## Re: integration

Maybe you can try with the t-formulas, let $t=\tan\left(\frac{x}{2}\right)$

3. ## Re: integration

integarte {secx}\{1+2
sin^2(x)},dx
(Quoted from mymathforum, where you cross-posted...)
We are given:

$\int\frac{\sec x}{1+2\sin^2x}\,dx$

Let's rewrite the integrand:

$\frac{\sec x}{1+2\sin^2x}=\frac{\cos x}{3}$$\frac{1+2\sin^2x+2\(1-\sin^2x$$}{\cos^2x$$1+2\sin^2x$$}\)=\frac{\cos x}{3}$$\frac{1}{1-\sin^2x}+\frac{2}{1+2\sin^2x}$$=$

$\frac{\cos x}{6}$$\frac{2}{1-\sin^2x}+\frac{4}{1+2\sin^2x}$$=\frac{\cos x}{6}$$\frac{\(\sin x-1$$-$$\sin x+1$$}{\sin^2x-1}+\frac{4}{1+2\sin^2x}\)=$

$\frac{1}{6}$$\frac{\cos x}{\sin x+1}-\frac{\cos x}{\sin x-1}+2\sqrt{2}\frac{\sqrt{2}\cos x}{\(\sqrt{2}\sin x$$^2+1}\)$

Now we may state:

$\int\frac{\sec x}{1+2\sin^2x}\,dx=\frac{1}{6}$$\ln|\sin x+1|-\ln|\sin x-1|+2\sqrt{2}\tan^{\small{-1}}\(\sqrt{2}\sin x$$\)+C$

Almost simultaneously, galactus wrote:

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

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

Let $t=\sqrt{2}sin(x), \;\ dt=\sqrt{2}cos(x)dx$

Notice now we have the form $\frac{1}{1+t^{2}}$

Continuing:

Since $t=\sqrt{2}sin(x)$, then $sec(x)=\sqrt{\frac{2}{2-t^{2}}}$ (this just comes from looking at a right triangle).

$dx=\frac{dt}{\sqrt{2}}sec(x)\Rightarrow dx=\frac{1}{\sqrt{2}}\sqrt{\frac{2}{2-t^{2}}}dt=\frac{1}{\sqrt{2-t^{2}}}dt$

Put all together:

$\int\frac{\sqrt{\frac{2}{2-t^{2}}}}{1+t^{2}}\cdot\frac{1}{2-t^{2}}dt=\frac{\sqrt{2}}{3}\int\frac{1}{t^{2}+1}dt +\frac{\sqrt{2}}{3}\int\frac{1}{2-t^{2}}dt$

The left most integral is an arctan. The right one is of the form $\int\frac{1}{a^{2}-t^{2}}dt=\frac{1}{2a}ln\left|\frac{t+a}{t-a}\right|$

So, we get after resubbing:

$\frac{ln|sin(x)+1|}{6}-\frac{ln|sin(x)-1|}{6}+\frac{\sqrt{2}}{3}tan^{-1}(\sqrt{2}sin(x))$

http://www.mymathforum.com/viewtopic...start=0#p89474

4. ## Re: integration

$\int\frac{\sec(x)}{1+2\sin^2(x)}dx=\int\frac{\cos( x)}{\cos^2(x)(1+2\sin^2(x))}dx$
$=\int\frac{\cos(x)}{(1-\sin^2(x))(1+2\sin^2(x))}dx$
substitute $\sin(x)=t$ to get
$\int\frac{1}{(1-t^2)(1+2t^2)}dx$
now partial fractions can be used to integrate

5. ## Re: integration

\displaystyle \begin{align*}\int{\frac{\sec{x}}{1 + 2\sin^2{x}}\,dx} &= \int{\frac{1}{\cos{x}(1 + 2\sin^2{x})}\,dx} \\ &= \int{\frac{\cos{x}}{\cos^2{x}(1 + 2\sin^2{x})}\,dx} \\ &= \int{\frac{\cos{x}}{(1 - \sin^2{x})(1+ 2\sin^2{x})}\,dx} \\ &= \int{\frac{1}{(1 - u^2)(1 + 2u^2)}\,du} \end{align*}

You can now integrate this using partial fractions.