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 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. 
