Question

If $x={\int }_0^y\frac{dt}{\sqrt{t-t^2}}$, then $\frac{dy}{dx}$ is equal to

• A. $y$
• B. $\sqrt{1+y^2}$
• C. $\frac{y}{\sqrt{1+y^2}}$
• D. $y^2$

Answer 1

The correct option is B $\sqrt{1+y^2}$

$\begin{array}{c}x={\int }_0^y\frac{dt}{\sqrt{t-t^2}}\\ x={\int }_0^y\frac{dt}{\sqrt{\frac{1}{4}-{\left(t-\frac{1}{2}\right)}^2}}\\ x={\left[2{\sin }^{-1}\left(2t-1\right)\right]}_0^4\\ x=2{\sin }^{-1}\left(2y-1\right)-3\pi \\ \frac{x+3\pi }{2}={\sin }^{-1}\left(2y-1\right)\\ y=\left(\sin \left(\frac{x}{2}+\frac{3\pi }{2}\right)+1\right)/2\\ y=\frac{-\cos \left(\frac{x}{2}\right)}{2}+\frac{1}{2}\\ dx=\frac{\sin \left(x/2\right)}{4}=\sqrt{1+y^2}\end{array}$