Evaluate `int_(0)^(pi)(x dx)/(1+cos alpha sin x)`,where `0lt alpha lt pi`.

Options

A. `(pialpha)/(sin alpha)`
B. `(pialpha)/(cosalpha)`
C. `(pialpha)/(1+sinalpha)`
D. `(pialpha)/(1+cosalpha)`

Answer

A

Explanation

Let `l = int_(0)^(pi)(x dx)/(1+cos alpha.sin x)` …(i)
`rArr l = int_(0)^(pi)((pi-x))/(1+cos alpha.sin(pi-x))dx` …(ii)
On adding Eqs. (i) and (ii), we get
`2l = pi int_(0)^(pi)(dx)/(1+cos alpha.sin x)=pi int_(0)^(pi)(dx)/(1+cos alpha((2tan x//2)/(1+tan^(2)x//2)))`
`= pi int_(0)^(pi)(sec^(2)x//2 dx)/((1+tan^(2)x//2)+cos alpha (2 tan x//2))`
Put `tan.(x)/(2)=t rArr ((1)/(2))sec^(2) x//2.dx`
`therefore " " 2l = pi int_(0)^(oo)(2dt)/(1+t^(2)+2tcos alpha)`
`l = pi int_(0)^(oo) (dt)/(1+t^(2)+2t cos alpha)`
`= pi int_(0)^(oo) (dt)/((t+cos alpha)^(2)+sin^(2)alpha)`
`= (pi)/(sin alpha)[tan^(-1)((t+cos alpha)/(sin alpha))]_(0)^(oo)`
`rArr l=(pi alpha)/(sin alpha)`