Question

The value of ${\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{{\sin }^{2}x+{\cos }^{2}x}dx$ is

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

${\int }_0^{\frac{\pi }{2}}\frac{si{n}^{2}x}{si{n}^{2}x+co{s}^{2}x}dx={\int }_0^{\frac{\pi }{2}}\frac{1}{1+co{t}^{2}x}dx$

$={\int }_0^{\frac{\pi }{2}}\frac{1}{cose{c}^{2}x}dx={\int }_0^{\frac{\pi }{2}}si{n}^{2}xdx=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}2si{n}^{2}xdx$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}(1-cos2x)dx=\frac{1}{2}[x-\frac{sin2x}{2}{]}_0^{\frac{\pi }{2}}$

$\frac{1}{2}(\frac{\pi }{2}-0)-\frac{1}{2}[\frac{sin\frac{2\pi }{2}}{2}-\frac{sin0}{2}]$

$=\frac{\pi }{4}$