Mathematics - Application of Derivative Question with Solution | TestHub

Let $\mathrm{x}={\cos }^2\theta$, so that $\mathrm{dx}=-2\mathrm{a}\sin \theta \cos \theta \mathrm{d}\theta$ Now ${\int }_0^a\sqrt{\frac{a-x}{x}}\mathrm{dx}={\int }_{\frac{\pi }{2}}^0\sqrt{\frac{a-a{\cos }^2\theta }{a{\cos }^2\theta }}$ $[\because \theta =\frac{\pi }{2}$ at $x=0;\theta =0$ at $x=a]$ $=a{\int }_0^{\pi /2}\sqrt{\frac{1-{\cos }^2\theta }{{\cos }^2\theta }}2\sin \theta \cos \theta d\theta$ $=a{\int }_0^{\pi /2}2\frac{\sin \theta }{\cos \theta }\cdot \sin \theta \cos \theta d\theta$ $=a{\int }_0^{\pi /2}2{\sin }^2\theta d\theta =a{\int }_0^{\pi /2}(1-\cos 2\theta )d\theta$ $\left[\because \cos 2\theta =1-2{\sin }^2\theta \right]$ $=\mathrm{a}{\left[\theta -\frac{\sin 2\theta }{2}\right]}_0^{\pi /2}$ $=\mathrm{a}\left[\left(\frac{\pi }{2}-\frac{\sin \pi }{2}\right)-(0-0)\right]$ $=\mathrm{a}\left(\frac{\pi }{2}-0\right)=\frac{\mathrm{a}\pi }{2}$