Another Trigonometric Integral

$I=\dfrac{4\cos x}{3+\red{12\sin^2x}-2\cos 2x+\red{12\sin^2x\cos 2x} + 3\cos 4x}$ $=\dfrac{4\cos x}{3+12\sin^2x(\red{1+\cos 2x})-2\blue{\cos 2x}+ 3\cos 4x}$ $=\dfrac{4\cos x}{3+24\sin^2x\cos^2x-2(\blue{2\cos^2x-1})+ 3\cos 4x}$ $=\dfrac{4\cos x}{\blue{3}+24\sin^2x\cos^2x-4\cos^2x+\blue{2}+ 3\red{\cos 4x}}$ $=\dfrac{4\cos x}{{5}+24\sin^2x\cos^2x-4\cos^2x+ 3(2\red{\cos^2 2x}-1)}$ $=\dfrac{4\cos x}{\blue{5}+24\sin^2x\cos^2x-4\cos^2x+ 6{(2\cos^2x-1)}^2\blue{-3}}$ $=\dfrac{4\cos x}{2+24\sin^2x\cos^2x-4\cos^2x+ 6{(4\cos^4x+1-4\cos^2x)}}$ $=\dfrac{2\cos x}{1+12\sin^2x\cos^2x-2\cos^2x+ 3(4\cos^4x+1-4\cos^2x)}$ $=\dfrac{2\cos x}{\blue{1}+\red{12\sin^2x\cos^2x}-2\cos^2x+ 12\cos^4x+\blue{3}-\red{12\cos^2x}}$ $=\dfrac{2\cos x}{4+12\cos^2x(\red{\sin^2x-1})-2\cos^2x+ 12\cos^4x}$ $=\dfrac{\blue{2}\cos x}{\blue{4}\cancel{\red{-12\cos^4x}}-\blue{2}\cos^2x+ \cancel{\red{12\cos^4x}}}$ $=\dfrac{\cos x}{2-\cos^2x}$ $\Rightarrow \int_0^{\frac{\pi}{2}}Idx=\int_0^{\frac{\pi}{2}}\dfrac{\red{\cos xdx}}{2-\red{\cos^2x}}$ $=\int_0^{\frac{\pi}{2}}\dfrac{d\sin x}{2-(1-\sin^2x)}$ $=\int_0^{\frac{\pi}{2}}\dfrac{d\sin x}{1+\sin^2x}$ $=[\arctan(\sin x)]_0^{\frac{\pi}{2}}$ $=\arctan(\sin \frac{\pi}{2})-\arctan(\sin 0)$ $=\red{\arctan 1}-\blue{\arctan 0}$ $=\dfrac{\pi}{4}-0=\red{\dfrac{1}{4}}\pi=\red{a}\pi$ $\Rightarrow \boxed{a=\dfrac{1}{4}=0.25}$

By expressing all the terms in the denominator in terms of $\cos(2x)$ , we can simplify the denominator of the integrand to $6 - 2\cos(2x) = 6 - 2(1 - 2\sin^2 x)$

Apply the substitution of $y = \sin x,$ the integral breaks down to $\frac\pi4 .$

The denominator is equivalent to $12\sin^2x\underbrace{(1+\cos(2x)}_{2\cos^2x}) + 3\underbrace{(1+\cos(4x)}_{2cos^2(2x)}) - 2\cos(2x) = 24\sin^2x\cos^2x + 6\cos^2(2x) - 2\cos(2x)$ $= 6(\underbrace{2\sin x\cos x}_{\sin(2x)})^2 + 6\cos^2(2x) - 2\cos(2x)$ $= 6\sin^2(2x) + 6\cos^2(2x) - 2\cos(2x)$ $= 6(\underbrace{\sin^2(2x) + 6\cos^2(2x)}_1) - 2\cos(2x)$ $= 6 - 2\cos(2x)$ Then, $\int_{x = 0}^{\frac{\pi}{2}} \frac{4\cos x}{6 - 2\cos(2x)} dx= \int_{x = 0}^{\frac{pi}{2}} \frac{2\cos x}{3 - (1 - 2\sin^2x)}dx$ $= \int_{x = 0}^{\frac{\pi}{2}} \frac{2\cos x}{2 + 2\sin^2x} dx$ $= \int_{x = 0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin^2x} dx$ Let $u = \sin x$ so $du = \cos x dx$ . $\int_{u = 0}^{1} \frac{1}{1+u^2} du = \arctan(1) - \arctan(0) = \boxed{\frac{\pi}{4}}$