` int_(0)^(pi//2) (cos^(5)x)/(sin^(5)x+cos^(5)x)dx` is equal to

Options

A. 0
B. `pi/3`
C. `pi/4`
D. `pi/2`

Explanation

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx, \]

we can use the property of definite integrals which states that

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In our case, \( a = 0 \) and \( b = \frac{\pi}{2} \). Thus, we can write:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5\left(\frac{\pi}{2} - x\right)}{\sin^5\left(\frac{\pi}{2} - x\right) + \cos^5\left(\frac{\pi}{2} - x\right)} \, dx. \]

Using the trigonometric identities \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \) and \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \), we can rewrite the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x}{\cos^5 x + \sin^5 x} \, dx. \]

Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx \)
2. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x}{\sin^5 x + \cos^5 x} \, dx \)

Adding these two equations gives:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\cos^5 x}{\sin^5 x + \cos^5 x} + \frac{\sin^5 x}{\sin^5 x + \cos^5 x} \right) \, dx. \]

The denominators are the same, so we can combine the numerators:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 x + \sin^5 x}{\sin^5 x + \cos^5 x} \, dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx. \]

Now we can evaluate the integral:

\[ 2I = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]

Thus, we find:

\[ I = \frac{\pi}{4}. \]

Therefore, the value of the integral is

\[ \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx = \frac{\pi}{4}. \]