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

The correct answer is 2.
$I=\frac{8}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{(\cos x{)}^{2023}}{(\sin x{)}^{2023}+(\cos x{)}^{2023}}dx$......(1)
Using $\underset{0}{\overset{a}{\int }}f(x)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$
$I=\frac{8}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{(\sin x{)}^{2023}}{(\sin x{)}^{2023}+(\cos x{)}^{2023}}dx$........(2)
Adding (1) & (2)
$2I=\frac{8}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}1dx$
$I=2$

Answer 2

Let: \[ I = \frac{8}{\pi} \int_{0}^{\pi/2} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx \quad \cdots (1). \] Using the property: \[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx, \] substitute \( a = \frac{\pi}{2} \) and \( f(x) = \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} \). Then: \[ I = \frac{8}{\pi} \int_{0}^{\pi/2} \frac{(\sin x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx \quad \cdots (2). \] Adding (1) and (2): \[ 2I = \frac{8}{\pi} \int_{0}^{\pi/2} 1 \, dx. \] Evaluate the integral: \[ 2I = \frac{8}{\pi} \cdot \frac{\pi}{2} = 4. \] Thus: \[ I = 2. \] Final Answer: 2.