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Prove that:`y=int_(1/8)^(sin^2x)sin^(-1)sqrt(t)dt+int_(1/8)^(cos^2x)cos^(-1),w h e r e0lt=xlt=pi/2,`is the equation of a straight line parallel to the x-axis. Find theequation. |
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Answer» 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` |
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