Question

Solve ${\int }_0^{\pi /2}\frac{{\cos }^{2}x}{{\sin }^{2}x+{\cos }^{2}x}dx$

• A. $-\frac{\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. None of these

Answer 1

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

Consider the given integral.

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{{\cos }^{2}x}{{\sin }^{2}x+{\cos }^{2}x}\right)dx$ …….. (1)

We know that

${\int }_b^{a}f\left(x\right)dx={\int }_b^{a}f(a+b-x)dx$

Therefore,

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{{\sin }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}\right)dx$ …… (2)

On adding equation (1) and (2), we get

$2I={\int }_0^{\frac{\pi }{2}}\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}\right)dx$

$2I={\int }_0^{\frac{\pi }{2}}1dx$

$2I={\left[x\right]}_0^{\frac{\pi }{2}}$

$2I=\frac{\pi }{2}-0$

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

Hence, this is the answer.