evalute int[ from (0) to( pi)]{1/t-cosx}dx t>0

Dear Baha, 

I= int(0 to pi){1/(t-cosx)}dx, let tan(x/2)=s, dx= 2ds/(1+s²),I=∫(0 to infinity)2ds/((1+s²)(t-cosx)==∫(0 to infinity)2ds/((1+s²)t+s²-1)=

(2/(t+1))∫(0 to infinity)ds/(s²+((t-1)/(t+1)))=(2/(t+1))[√((t+1)/(t-1))(arctan(s√((t+1)/(t-1))¦_0^infinity = π/√(t²-1)

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