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)

$\frac{\pi }{4}$

Answer 2

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