Question

${\int }_0^{\pi /2}\frac{{\sin }^{100}x}{{\sin }^{100}x+{\cos }^{100}x}dx$ is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{8}$

Answer 1

Answer: (C)

${\int }_0^{\pi /2}\frac{{\sin }^{100}x}{{\sin }^{100}x+{\cos }^{100}x}dx$
$={\int }_0^{\pi /2}\frac{{\sin }^{100}}{{\sin }^{100}x+{\sin }^{100}\left(\frac{\pi }{2}-x\right)}dx$
$=\frac{\frac{\pi }{2}-0}{2}=\frac{\pi }{4}$
$\left[\because {\int }_a^b\frac{f(x)dx}{f(x)+f(a+b-x)}=\frac{b-a}{2}\right]$
Alternate $I={\int }_0^{\pi /2}\frac{{\sin }^{100}x}{{\sin }^{100}x+{\cos }^{100}x}dx$ .. (i)
$I={\int }_0^{\pi /2}\frac{{\sin }^{100}\left(\frac{\pi }{2}-x\right)}{{\sin }^{100}\left(\frac{\pi }{2}-x\right)-{\cos }^{100}\left(\frac{\pi }{2}-x\right)}dx$
$\left[\begin{array}{rl}& \because bydefine\text{integral property}\text{.}\\ & {\int }_a^{0}f(x)dx={\int }_0^{a}f(a-x)dx\end{array}\right]$
$I={\int }_0^{\pi /2}\frac{{\cos }^{100}x}{{\cos }^{100}x+{\sin }^{100}x}dx$ .. (ii)
On adding Eqs. (i) and (ii), we get
$2I={\int }_0^{\pi /2}1dx=\frac{\pi }{2}-0\Rightarrow I=\frac{\pi }{4}$