Answer
$$\frac{1}{\pi }\ln \left| {\tan \pi x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sin \pi x\cos \pi x}}dx} \cr
& {\text{Let }}u = \pi x,{\text{ }}du = \pi dx,{\text{ }}dx = \frac{1}{\pi }du \cr
& {\text{Substituting}} \cr
& \int {\frac{1}{{\sin \pi x\cos \pi x}}dx} = \int {\frac{1}{{\sin u\cos u}}\left( {\frac{1}{\pi }} \right)du} \cr
& = \frac{1}{\pi }\int {\frac{1}{{\sin u\cos u}}du} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{1}{{\sin u\cos u}}du = \ln \left| {\tan u} \right| + C,{\text{ then}}} \cr
& = \frac{1}{\pi }\ln \left| {\tan u} \right| + C \cr
& {\text{Write in terms of }}u,{\text{ let }}u = \pi x \cr
& = \frac{1}{\pi }\ln \left| {\tan \pi x} \right| + C \cr} $$
