Mathematics - Definite Integration Question with Solution | TestHub

$I=2{\int }_0^{\frac{\pi }{2}}\underset{I_1}{\underbrace{{\sin }^{2}x\cdot \sqrt{\frac{\pi x}{2}-x^2}}}-{\int }_0^{\frac{\pi }{2}}g(x)dx$ Let $I_1={\int }_0^{\frac{\pi }{2}}{\sin }^{2}x\sqrt{{\left(\frac{\pi }{4}\right)}^2-{\left(x-\frac{\pi }{4}\right)}^2}$ (making perfect square) apply kings $I_1={\int }_0^{\frac{\pi }{2}}{\cos }^{2}x\sqrt{{\left(\frac{\pi }{4}\right)}^2-{\left(\frac{\pi }{2}-x\right)}^2}$ add both $2I_1={\int }_0^{\frac{\pi }{2}}\sqrt{{\left(\frac{\pi }{4}\right)}^2-{\left(x-\frac{\pi }{4}\right)}^2}$ i.e. $2I_1={\int }_0^{\frac{\pi }{2}}g(x)$ Now $I=2I_1-{\int }_0^{\frac{\pi }{2}}g(x)=0$