### Understanding the Problem The integral in question is: \[ \int_{0}^{\pi/2} \frac{\sqrt{\cos^8 x} \, dx}{\sqrt{\cos^8 x + \sqrt{\sin^8 x}}} = \frac{\pi}{4} \] At first glance, the integrand looks complex due to the nested square roots and trigonometric functions raised to the eighth power. My goal is to simplify this expression to make the integral more manageable. ### Simplifying the Integrand Let's break down the integrand: \[ \frac{\sqrt{\cos^8 x}}{\sqrt{\cos^8 x + \sqrt{\sin^8 x}}} \] First, simplify the square roots: 1. \(\sqrt{\cos^8 x} = (\cos^8 x)^{1/2} = \cos^4 x\), since \(\cos x\) is non-negative in the interval \([0, \pi/2]\). 2. \(\sqrt{\sin^8 x} = (\sin^8 x)^{1/2} = \sin^4 x\), similarly because \(\sin x\) is non-negative in \([0, \pi/2]\). So, the integrand simplifies to: \[ \frac{\cos^4 x}{\sqrt{\cos^8 x + \sin^4 x}} \] Wait a minute—this seems incorrect. Let's re-examine the original expression. The denominator is \(\sqrt{\cos^8 x + \sqrt{\sin^8 x}}\), and we've simplified \(\sqrt{\sin^8 x}\) to \(\sin^4 x\). So, the correct simplified integrand is: \[ \frac{\cos^4 x}{\sqrt{\cos^8 x + \sin^4 x}} \] But the original equation claims that this integral equals \(\pi/4\). To verify this, I need to evaluate the integral numerically or find a clever substitution or symmetry. ### Attempting a Substitution Given the limits from \(0\) to \(\pi/2\), a common strategy is to use the substitution \(x = \pi/2 - t\). This often helps in exploiting symmetries between sine and cosine functions. Let \(x = \pi/2 - t\), then \(dx = -dt\), and the limits change as follows: - When \(x = 0\), \(t = \pi/2\). - When \(x = \pi/2\), \(t = 0\). The integral becomes: \[ I = \int_{\pi/2}^{0} \frac{\cos^4 (\pi/2 - t)}{\sqrt{\cos^8 (\pi/2 - t) + \sin^4 (\pi/2 - t)}} (-dt) = \int_{0}^{\pi/2} \frac{\sin^4 t}{\sqrt{\sin^8 t + \cos^4 t}} dt \] Here, I used \(\cos(\pi/2 - t) = \sin t\) and \(\sin(\pi/2 - t) = \cos t\). Now, the original integral \(I\) is: \[ I = \int_{0}^{\pi/2} \frac{\cos^4 x}{\sqrt{\cos^8 x + \sin^4 x}} dx \] And after substitution, we have: \[ I = \int_{0}^{\pi/2} \frac{\sin^4 x}{\sqrt{\sin^8 x + \cos^4 x}} dx \] Adding these two expressions for \(I\): \[ 2I = \int_{0}^{\pi/2} \left( \frac{\cos^4 x}{\sqrt{\cos^8 x + \sin^4 x}} + \frac{\sin^4 x}{\sqrt{\sin^8 x + \cos^4 x}} \right) dx \] This seems promising. Let's see if the sum inside the integral simplifies. ### Combining the Integrands Consider the sum: \[ \frac{\cos^4 x}{\sqrt{\cos^8 x + \sin^4 x}} + \frac{\sin^4 x}{\sqrt{\sin^8 x + \cos^4 x}} \] Let’s denote \(a = \cos^4 x\) and \(b = \sin^4 x\). Then the sum becomes: \[ \frac{a}{\sqrt{a^2 + b}} + \frac{b}{\sqrt{b^2 + a}} \] This seems symmetric but not immediately simplifiable. Maybe rationalizing or finding a common