IIT JEE , AIEEE Forum - Physics , General - sawaal no 4...............................................................................

ramkumar_november

the given integral is

I = $int_0^{pi} rac{x;;;;dx}{1-cosalpha sinx}$ .........(1)

applying property 4 ....

I = $\int_0^{\pi}\frac{(\pi-x)dx}{1-cos\alpha sinx}$ ..........(2)

adding (1) and (2) ...

2I = $pi;;int_0^{pi} rac{dx}{1-cosalpha sinx}$

now put tan x/2 = t and then simplifying we get the final integral as : :

I = $rac{pi}{2};;int_0^{infty} rac{2dt}{t^2-2tcosalpha+1}$

I = $pi;;int_0^{infty} rac{dt}{sinalpha + (t-cosalpha)^2}$

I = $rac{pi}{sinalpha};;Bigg[,tan^{-1}Bigg( rac{t-cosalpha}{sinalpha}Bigg)Bigg]_0^{infty}$

I =

This is the final answer........