Integration of $1/(1+\sin 2x)$
Okay so I've to integrate $$∫\frac{dx}{(1+\sin 2x)}$$ but I'm not getting any further from here $$∫\frac{dx}{(\sin x+\cos x)^2}.$$ Any hints or suggestions?
$\endgroup$ 05 Answers
$\begingroup$Let $$I = \int\frac{1}{1+\sin 2x}dx $$
Put $\displaystyle 2x=\frac{\pi}{2}-2t\;,$ Then $dx = -dt$
So $$I = -\int\frac{1}{1+\cos 2t}dt = -\int\frac{1}{2\cos^2 t}dt = -\frac{1}{2}\int \sec^2 tdt = -\frac{1}{2}\tan t+\mathcal{C}$$
So $$I = \int\frac{1}{1+\sin 2x}dx=-\frac{1}{2}\tan \left(\frac{\pi}{4}-x\right)+\mathcal{C}$$
$\endgroup$ 2 $\begingroup$$\displaystyle\int\frac{1}{1+\sin 2x}dx=\int\frac{1}{1+\sin 2x}\cdot\frac{1-\sin 2x}{1-\sin 2x}dx=\int\frac{1-\sin 2x}{\cos^2 2x}dx$
$\displaystyle=\int\left(\sec^2 2x-\sec 2x\tan 2x\right)dx=\frac{1}{2}\big(\tan 2x-\sec 2x\big)+C$
$\endgroup$ 2 $\begingroup$see my nice answer using OP's step:
$$\int \frac{dx}{1+\sin 2x} =\int \frac{dx}{(\sin x+\cos x)^2}$$$$=\int \frac{dx}{\cos^2 x(1+\tan x)^2}=\int \frac{\sec^2 xdx}{(1+\tan x)^2}$$ $$=\int \frac{d(1+\tan x)}{(1+\tan x)^2}=-\frac{1}{1+\tan x}+C$$
$\endgroup$ 2 $\begingroup$An other way
Let $x=\arctan u$, then $\sin(2x)=\frac{2u}{1+u^2}$, $$...=\int^{\tan(x)}\frac{1+u^2}{1+u^2+2u}\mathrm d u=\int^{\tan(x)}\frac{1+u^2}{(1+u)^2}\mathrm d u.$$ Use the fact that $1+u^2=(1+u)^2-2u$ and conclude.
$\endgroup$ $\begingroup$Alternatively, from where you left off, the integrand can be written as $$\frac 12\csc^2(x+\frac{\pi}{4})$$ so you end up with $$-\frac 12\cot(x+\frac{\pi}{4})+c$$ which is equivalent to @juantheron's answer
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