The range of the function `f(x)=int_(-1)^(1)(sinxdt)/(1+2tcosx+t^(2))` is

Options

A. `[-(pi)/2,(pi)/2]`
B. `[0,pi]`
C. `{0,pi}`
D. `{-(pi)/2,(pi)/2}`

Answer

D

Explanation

We have
`f(x)=int_(-1)^(1)(sinx dt)/(sin^(2)x+(t-cosx)^(2))`
`=(sinx)/(sinx)tan^(-1)((t-cosx)/(sinx))|_(-1)^(1)`
`=tan^(-1)((1-cosx)/(sinx))-tan^(-1)((-1-cosx)/(sinx))`
`=tan^(-1) (tanx//2) +tan^(-1) (cotx//2)`
Now we know that `tan^(-1)x+tan^(-1)1/x={((pi)/2,xgt0),(-(pi)/2,xlt0):}`
or `tan^(-1)("tan"x/2)+tan^(-1)(1/("tan"x/2))={((pi)/2, "tan"(pi)/2gt0),(-(pi)/2,"tan"x/2lt0):}`
Hence range of `f(x)` is `{-(pi)/2,(pi)/2}`