solving Integration of trignometry In the following integral
$$\int \frac {1}{\sec x+ \mathrm {cosec} x} dx $$
My try: Multiplied and divided by $\cos x$ and Substituting $\sin x =t$. But by this got no result.
 A: A big partial fraction decomposition!!:
We use Weierstrass substitution and use $u =\tan \frac {x}{2} $ to get $$I =\int \frac {1}{\sec x+\mathrm {cosec}x } dx =\int \frac {\sin x\cos x}{\sin x+\cos x} dx= 4\int \frac {u (u^2-1)}{(u^2+1)^2 (u^2-2u-1)} du =4I_1 $$ Now we have  $$I_1 =\int \frac {u (u^2-1)}{(u^2+1)^2 (u-\sqrt {2}-1)(u+\sqrt {2}-1)} = -\frac {1}{4} \int \frac {1}{u^2+1} du +\frac{1}{2} \int \frac{u+1}{(u^2+1)^2} du +\frac {4-3\sqrt {2}}{48-32\sqrt {2}} \int \frac {1}{u+\sqrt {2}-1} du +\frac {4+3\sqrt {2}}{32\sqrt {2}+48} \int \frac {1}{u-\sqrt {2}-1} du =I_{11} +I_{12} +I_{13} +I_{14} $$ Hope you can take it from here. 
A: Since I don't know LaTex, I am going to add a link so that you can understand with your best of abilities. Let me know if you find a problem/bug.
Since, I could only come up with a link. Let me know if there is any problem in it if you come across. I'll always have your back ! 
A: $\displaystyle I = \int\frac{1}{\sec x+\csc x}dx = \frac{1}{2}\int\frac{\sin 2x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{(\sin x+\cos x)^2-1}{\sin x+\cos x}dx$
$\displaystyle I = \frac{1}{2}\int (\sin x+\cos x)dx - \frac{1}{2}\int \frac{1}{\sin x+\cos x}dx$
$\displaystyle I =\frac{1}{2}(\sin x-\cos x)-\frac{1}{2\sqrt{2}}\int \frac{1}{\sin (x+45^0)}dx$
$\displaystyle I = \frac{1}{2}(\sin x-\cos x)-\frac{1}{2\sqrt{2}}\int \csc (x+45^0)dx$
Use $\displaystyle \int \csc  xdx = \ln |\csc x-\cot x|+\mathcal{C} = \ln\tan \frac{x}{2}+\mathcal{C}$
A: HINT 
Use half-angle formulae to substitute for $\sec x$ and $\csc x$, and use the substitution $u = \tan \frac{x}{2}$, to get the following expression:
$$I = 4\int\frac{u(u^2-1)}{(u^2+1)^2(u^2-2u-1)}du$$
This should be easier to compute, using factorization and other basic techniques
