Symmetry Property and King's Property in Definite Integrals
The given integral is:
\[ \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(4 + \sin x)^2} \, dx \]
A suitable method to solve this integral is to use a symmetry property known as the King's property in definite integrals, which uses the substitution \( x = \frac{\pi}{2} - t \) to exploit the symmetrical limits of integration from 0 to \( \frac{\pi}{2} \).
Let's apply the King's property and see what we get:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2} - t)}{(4 + \sin(\frac{\pi}{2} - t))^2} \, dt \]
We know that \( \cos(\frac{\pi}{2} - t) = \sin t \) and \( \sin(\frac{\pi}{2} - t) = \cos t \), so substituting these in, we get:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{(4 + \cos t)^2} \, dt \]
If we now add the original integral and the transformed integral, we get:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\cos x}{(4 + \sin x)^2} + \frac{\sin x}{(4 + \cos x)^2} \right) dx \]
Now, since the integrand is symmetric about the line \( y = x \), we can rewrite the integral in terms of a single variable:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x + \sin x}{(4 + \sin x)^2} \, dx \]
By observation or using trigonometric identities, notice that:
\[ 4 + \sin x + \cos x = (4 + \sin x)(1 + \sec x) \]
Now differentiate the denominator with respect to x.
\[ d(4+\sin x) = \cos x \,dx \]
Now,
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{d(4 + \sin x)}{(4 + \sin x)^2} \]
This can now be easily integrated using the formula for the integral of a derivative of a function over the function squared:
\[ 2I = \left. -\frac{1}{4 + \sin x} \right|_{0}^{\frac{\pi}{2}} \]
When we plug in the limits, we get:
\[ 2I = \left( -\frac{1}{4 + \sin(\frac{\pi}{2})} \right) - \left( -\frac{1}{4 + \sin(0)} \right) \]
\[ 2I = \left( -\frac{1}{5} \right) - \left( -\frac{1}{4} \right) \]
\[ 2I = -\frac{1}{5} + \frac{1}{4} \]
\[ 2I = \frac{1}{4} - \frac{1}{5} \]
\[ 2I = \frac{5 - 4}{20} \]
\[ 2I = \frac{1}{20} \]
Hence, the value of the original integral is:
\[ I = \frac{1}{2} \cdot \frac{1}{20} = \frac{1}{40} \]
