Evaluate the following :
`int_(0)^(pi//2)(1)/(5+4 sinx)dx`

Explanation

Let `I=int_(0)^(pi//2)(1)/(5+4sinx)dx`
Put `tan.(x)/(2)=t " "therefore" "x=2 tan^(-1)t`
`therefore dx=(2dt)/(1+t^(2))and sin x=(2t)/(1+t^(2))`
When `x=0, t= tan 0=0" When "x=(pi)/(2), t= tan.(pi)/(4)=1`
`therefore I=int_(0)^(1)(1)/(5+4((2t)/(1+t^(2)))).(2dt)/(1+t^(2))`
`=int_(0)^(1)(1+t^(2))/(5+5t^(2)+8t).(2dt)/(1+t^(2))`
`=2int_(0)^(1)(1)/(5t^(2)+8t+5)dt`
`=(2)/(5)int_(0)^(1)(1)/(t^(2)+(8)/(5)t+1)dt`
`=(2)/(5)int_(0)^(1)(1)/((t^(2)+(8)/(5)t+(16)/(25))-(16)/(25)+1)dt`
`=(2)/(5)int_(0)^(1)(1)/(t+(4)/(5))^(2)+((3)/(5))^(2)dt`
`=(2)/(5)xx(1)/((3//5))[tan^(-1)((t+4//5)/(3//5))]_(0)^(1)`
`=(2)/(3)[tan^(-1)((5t+4)/(3))]_(0)^(1)`
`(2)/(3)[tan^(-1)3-tan^(-1).(4)/(3)]`
`=(2)/(3)tan^(-1)((3-4//3)/(1+3(4//3)))=(2)/(3)tan^(-1)((9-4)/(3+12))`
`=(2)/(3)tan^(-1)((5)/(15))=(2)/(3)tan^(-1)((1)/(3)).`