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

Answer

`(pi alpha)/(sin alpha)`

Explanation

Let `I=int_(0)^(pi)(xdx)/(1+cos alpha sinx )`…………..1
`=int_(0)^(pi)((pi-x)dx)/(1+cos alpha (sin (pi-x)))`
[using `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`:.I=int_(0)^(pi)((pi-x)dx)/(1+cos alpha sinx)`…………2
Adding 1 and 2 we get
`2I=int_(0)^(pi)(x+pi-x)/(1+cos alpha sinx) dx`
`=int_(0)^(pi)(pi)/(1+cos alpha sin x)dx`
`:.I=(pi)/2 int_(0)^(pi)1/(1+cos alpha sin x)dx`
`=(pi)/2 int_(0)^(pi)1/(1+cos alpha xx(2tanx//2)/(1+tan^(2)x//2) dx`
`=(pi)/2 int_(0)^(pi)(sec^(2)x//2)/(1+tan^(2)x//2+2cos alpha tan x//2)`
Put `tanx//2=t` or `1/2"sec^(2)x/2dx=dt`
Also, when `xto 0, t to 0`
and when `x to pi,t to oo`
`:.I=(pi)/2 int_(0)^(oo) (2dt)/(t^(2)+(2cos alpha)t+1)`
`=piint_(0)^(oo) (dt)/((t+cos alpha)^(2)+1-cos^(2)alpha`
`=pi . 1/(sin alpha)[tan^(-1)((t+cos alpha)/(sin alpha))]_(0)^(oo)`
`=(pi)/(sin alpha)[tan^(-1)oo-tan^(-)(cot alpha)]`
`=(pi)/(sin alpha)[(pi)/2-tan^(-1)(cot alpha)]`
`=(pi)/(sin alpha) [cot^(-1)(cot alpha)]`
`=(pi alpha)/(sin alpha)`