Question

The value of $\frac{8}{\pi} \int\limits_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} d x$ is

Answer 1

Correct Answer: 2

To solve the integral $\frac{8}{\pi} \int\limits_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} d x$, we observe that it involves symmetry properties of trigonometric functions. By the substitution $x = \frac{\pi}{2} - u$, with $dx = -du$, the limits change from $0$ to $\frac{\pi}{2}$ to $\frac{\pi}{2}$ to $0$. Upon substituting, we get:

$I = \int\limits_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} d x = \int\limits_0^{\frac{\pi}{2}} \frac{(\sin u)^{2023}}{(\cos u)^{2023}+(\sin u)^{2023}} (-du)$

Reversing the limits introduces a negative sign:

$I = \int\limits_0^{\frac{\pi}{2}} \frac{(\sin u)^{2023}}{(\cos u)^{2023}+(\sin u)^{2023}} du$

Now, add the original integral to its transformed version:

$$2I = \int\limits_0^{\frac{\pi}{2}} \left(\frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} + \frac{(\sin x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}\right) d x$$

$$2I = \int\limits_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}$$

This simplifies to:

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

Thus, the original integral evaluates to $\frac{\pi}{4}$. Multiplying by $\frac{8}{\pi}$ as per the problem statement, we find:

$\frac{8}{\pi} \times \frac{\pi}{4} = 2$

Therefore, the value indeed falls within the range of 2 to 2.