Question

${\int }_0^{\pi /2}\frac{{\sin }^{1000}xdx}{{\sin }^{1000}x+{\cos }^{1000}x}$ is equal to

• A. $1000$
• B. $1$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

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

Let $I={\int }_0^{\pi /2}\frac{{\sin }^{1000}xdx}{{\sin }^{1000}x+{\cos }^{1000}x}$.......(1)

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

$I={\int }_0^{\pi /2}\frac{{\sin }^{1000}(\frac{\pi }{2}-x)dx}{{\sin }^{1000}(\frac{\pi }{2}-x)+{\cos }^{1000}(\frac{\pi }{2}-x)}$

$\Rightarrow I={\int }_0^{\pi /2}\frac{{\cos }^{1000}xdx}{{\cos }^{1000}x+{\sin }^{1000}x}$.......(2)

Add (1) and (2)

$2I={\int }_0^{\pi /2}\frac{{\sin }^{1000}x+{\cos }^{1000}x}{{\sin }^{1000}x+{\cos }^{1000}x}dx={\int }_0^{\pi /2}dx$

$\Rightarrow 2I=[x{]}_0^{\pi /2}=\frac{\pi }{2}-0=\frac{\pi }{2}$

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