Mathematics - Definite Integration Question with Solution | TestHub

Now $I_1={\int }_0^{\frac{\pi }{2}}f(x)\cdot g(x)dx=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}g(x)dx$ (it is explained in previous question solution) i.e. $\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\sqrt{{\left(\frac{\pi }{4}\right)}^2-{\left(x-\frac{\pi }{4}\right)}^2}\mathrm{dx}$ Using $\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2{\sin }^{-1}\left(\frac{x}{a}\right)\right)+C$ $\Rightarrow \frac{1}{2}{\left[\frac{\left(x-\frac{\pi }{4}\right)}{2}\sqrt{\frac{\pi x}{2}-x^2}+\frac{\frac{{\pi }^2}{2}}{2}{\sin }^{-1}\left(\frac{x-\frac{\pi }{4}}{\frac{\pi }{4}}\right)\right]}_0^{\pi /2}$ $\begin{array}{rl}& \Rightarrow \frac{1}{2}\left[\left(0+\frac{{\pi }^3}{64}\right)-\left(0+\left(\frac{-{\pi }^3}{64}\right)\right)\right]\\ & \Rightarrow \frac{1}{2}\times \frac{{\pi }^3}{32}\end{array}$ Now $\frac{16}{{\pi }^3}\times \frac{{\pi }^3}{64}=\frac{1}{4}=0.25$