Prove that: `y=int_(1/8)^(sin^2x)sin^(-1)sqrt(t)dt+int_(1/8)^(cos^2x)cos^(-1)sqrt(t) dt`, where `0lt=xlt=pi/2`, is the equation of a straight line parallel to the x-axis. Find the equation.

Explanation

Here we have to prove that `y=` constant or derivative of `y` w.r.t `x` is zero. ltbr. `y=int_(1/8)^(sin^(2)x) sin^(-1)sqrt(5)dt, dt+int_(1/8)^(cos^(2)x) cos^(-1)sqrt(5)dt`…………..1
`(dy)/(dx)=sin^(-1)sqrt(sin^(2)x)2sinx cosx`
`+cos^(-1)sqrt(cos^(2)x)(-2cosxsinx)`
`=2x sin x cos x-2x sin x cos x`
`=0` for all `x`
Therefore the curve in equation 1 is a straight line parallel to the `x`-axis.
Now, since `y` is constant it is independent of `x`. So let us select `x=pi//4`.Then
`y=int_(1//8)^(1//2)sin^(-1)sqrt(5)dt+int_(1//8)^(1//2)cos^(-1)sqrt(t)dt`
`=int_(1//8)^(1//2)(sin^(-1)sqrt(5)+cos^(-1)sqrt(t))dt`
`=int_(1//8)^(1//2)pi//2dt`
`=(pi)/2[1/2-1/8]`
`=(3pi)/16`
Therefore, equation of the line is `y=(3pi)/16`