If `f(x)=int_(0)^(pi)(t sin t dt)/(sqrt(1+tan^(2)xsin^(2)t))` for `0lt xlt (pi)/2` then

Options

A. `f(0^(+))=-pi`
B. `f((pi)/4)=(pi^(2))/8`
C. `f` is continuous and differntiable in `(0,(pi)/2)`
D. `f` is continuous but not differentiable in `(0,(pi)/2)`

Answer

C

Explanation

`f(x)=int_(0)^(pi)(tsint)/(sqrt(1+tan^(2)x sin^(2)t)) dt `
Replacing `t` by `pi-t` and then adding `f(x)` with 1 we get
`f(x)=(pi)/2 int_(0)^(pi)(sint)/(sqrt(1+tan^(2)xsin^(2)t))dt`
`=pi int_(0)^(pi//2) (sint)/(sqrt(1+tan^(2)x(1-cos^(2)t))dt`
Let `y=cost`
`:.dy=-sint dt`
`:. f(x)=pi int_(0)^(1)(dy)/(sqrt(sec^(2)x-(tan^(2)x)y^(2)))`
`=(pi)/(tanx)int_(0)^(1)(dy)/(sqrt(cosec^(2)x-y^(2)))`
`=(pi)/(tanx){"sin"^(-1)y/(cosecx)}_(0)^(1)`
`=(pi)/(tanx) "sin^(-1)(sinx)=(pix)/(tanx)`