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

Options

A. `(pi alpha)/(sin alpha)`
B. `( pi alpha)/(cos alpha)`
C. `(sin alpha)/(1+sin alpha)`
D. `(pi alpha)/(1+cos alpha)`

Answer

A

Explanation

Let `=int _(0)^(pi)(x dx)/(1+cos alpha*sin x)" "...(i)`
`impliesI =int _(0)^(pi)=(dx)/(1+cos alpha *sin (pi-x))dx" "...(ii)`
On adding Eqs. (i) and (ii), we get
`therefore 2I =piint _(0)^(pi)(dx)/(1+cos alpha *sin x)`
`=piint _(0)^(pi)(dx)/(1+cos alpha ((2tan x//2)/(1+tan ^(2)x//2)))`
`=pi int _(0)^(pi)(sec ^(2)x//2dx)/((a+tan ^(2)x//2)+cos alpha (2 tan x//2))`
Put ` tan x//2=t`
`implies(1//2)sec ^(2)x//2dx=dt`
`therefore2I=piint _(0)^(oo)(2dt)/(1+t^(2)+2tcos alpha )`
`I=piint _(0)^(oo)(2dt)/(1+t^(2)+2t cos alpha )`
`=piint_(0)^(oo)(dt)/((t+cos alpha)^(2)+sin ^(2)alpha)`
`=(pi)/(sin alpha )"["tan ^(-1)((t+cos alpha )/(sin alpha ))"]"_(0)^(oo)`
`I=(pialpha )/(sin alpha )`