Question - Symmetry Property and King's Property in Definite Integrals

Solution:

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:\[ I = \frac{1}{2} \cdot \frac{1}{20} = \frac{1}{40} \]

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