integration
I got this problem in my homework exercise:
$$\int \arctan(\sec x + \tan x) dx$$
I simplified it to
$$\int \arctan\left(\dfrac{1+\sin x}{\cos x}\right) dx$$
$$=\int \arctan\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right) dx$$
now, I tried putting $\sqrt{\dfrac{1+\sin x}{1-\sin x}} = \tan t$
Then $x = \arcsin(\sin^2 t-\cos^2 t)$ but it becomes complete mess after that!
$\alpha = \cot^{-1} x\\ \beta = \tan^{-1} (\sin \alpha)$
Use trig indentities to find $\csc \alpha$ and $\sin \alpha$
$\sin\alpha = \tan \beta$
Use similar identities to find $\sec \beta$ and $\cos\beta$
You made a mistake : \begin{aligned}\int{\frac{\mathrm{d}x}{1-3\cos^{2}{x}\sin^{2}{x}}}&=\int{\frac{\sec^{2}{x}}{\sec^{2}{x}-3\color{red}{\sin^{2}{x}}}\,\mathrm{d}x}\\ &=\int{\frac{\mathrm{d}y}{1+y^{2}-3\left(1-\frac{1}{1+y^{2}}\right)}}\\ &=\int{\frac{y^{2}+1}{y^{4}-y^{2}+1}\,\mathrm{d}y}\\ &=\int{\frac{1+\frac{1}{y^{2}}}{\left(y-\frac{1}{y}\right)^{2}+1}\,\mathrm{d}y}\\ &=\int{\frac{\mathrm{d}u}{u^{2}+1}}\\ &=\arctan{u}+C\\ &=\arctan{\left(y-\frac{1}{y}\right)}+C\\ \int{\frac{\mathrm{d}x}{1-3\cos^{2}{x}\sin^{2}{x}}}&=\arctan{\left(\tan{x}-\cot{x}\right)}+C\end{aligned}
Best Answer
$${1+\sin(x)\over \cos(x)}={\sin(x/2)+\cos(x/2)\over \cos(x/2)-\sin(x/2)}={1+\tan(x/2)\over 1-\tan(x/2)}=\tan(\pi/4+x/2)$$
Also keep in mind: $$\tan^{-1}(\tan(z))=\begin{cases}z&-\pi/2\le z\le\pi/2\\ z-\pi&\ \ \ \pi/2\lt z\le \pi\\z+\pi&\ \ -\pi\le z\lt-\pi/2 \end{cases}$$