Question

Solve $\int \frac{dx}{\sin x+\sec x}$

Answer 1

Consider the following integral.

$I=\int \frac{1}{\sin x+\sec x}dx$

$=\int \frac{2\cos x}{2+2\sin x\cos x}dx$

$=\int \frac{\cos x-\sin x+\sin x+\cos x}{1+{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}dx$

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

$I=I_1+I_2$

Integrate $I_1$and $I_2$ we get,

$I_1=\int \frac{\sin x+\cos x}{1+{(\sin x+\cos )}^2}dx$$

Let, $t=\sin x+\cos x$

And differentiate both side w.r.t x

$dt=(\cos x-\sin x)dx$

$=\int \frac{\cos x-\sin x}{1+{\left(t\right)}^2}\ast \frac{1}{(\cos x-\sin x)}dt$

$=\int \frac{1}{1+{\left(t\right)}^2}dt$

$={\tan }^{-1}t+C$

$+{\tan }^{-1}(\sin x+\cos x)+C$

$I_2=\int \frac{\sin x+\cos x}{2+\sin 2x}dx$

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

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

$=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)}dx$

$u=\sin x-\cos x$

$dx=\frac{du}{\sin x+\cos x}$

$I_2=\int \frac{\sin x+\cos x}{\sqrt{3}-u^2}\frac{du}{(\sin x+\cos x)}$

$=\int \frac{1}{{\left(\sqrt{3}\right)}^2-u^2}du$

$=\frac{1}{\sqrt{3}}\ln \left(\frac{\sqrt{3}+u}{\sqrt{3}-u}\right)$

$=\frac{1}{\sqrt{3}}\ln \left(\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-(\sin x-\cos x)}\right)$

$I=\int \frac{1}{\sin x+\sec x}dx=\frac{1}{\sqrt{3}}\ln \left(\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-(\sin x-\cos x)}\right)+{\tan }^{-1}(\sin x+\cos x)+C$

Hence, this is the correct answer.