Sine and Cosine in integral

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {\cos^4 x + \sin x \cos^3 x + \sin^2 x \cos^2 x + \sin^3 x \cos x}{\cos^4 x + \sin^4 x + 2 \sin x \cos^3 x + 2 \sin^2 x \cos^2 x + 2 \sin^3 x \cos x} dx & \small \blue{\text{By }\int_a^b f(x) dx = \int_a^b f(a+b-x) dx} \end{aligned}$

$\small \begin{aligned} \ \ \ & = \frac 12 \int_0^\frac \pi 2 \frac {\cos^4 x + \sin x \cos^3 x + \sin^2 x \cos^2 x + \sin^3 x \cos x}{\cos^4 x + \sin^4 x + 2 \sin x \cos^3 x + 2 \sin^2 x \cos^2 x + 2 \sin^3 x \cos x} + \frac {\sin^4 x + \cos x \sin^3 x + \cos^2 x \sin^2 x + \cos^3 x \sin x}{\cos^4 x + \sin^4 x + 2 \sin x \cos^3 x + 2 \sin^2 x \cos^2 x + 2 \sin^3 x \cos x} dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac {\cos^4 x + \sin^4 x + 2 \sin x \cos^3 x + 2 \sin^2 x \cos^2 x + 2 \sin^3 x \cos x} {\cos^4 x + \sin^4 x + 2 \sin x \cos^3 x + 2 \sin^2 x \cos^2 x + 2 \sin^3 x \cos x} dx \\ & = \int_0^\frac \pi 2 dx = \boxed{\frac \pi 4} \end{aligned}$

Above integral can be written as $\textrm{Apple 🍎🍏}$

$I=\int_0^{\frac{π}{2}} \frac{cos^2 x (cos^2 x +sin^2 x)+sinx cosx (sin^2 x +cos^2 x)}{(sin^2 x+cos^2 x)^2 +2sinxcosx(sin^2 x+cos^2 x)}dx$

$I= \int_0^{\frac{π}{2}} \frac{cos^2 x +sinx cosx }{1+2 sinx cosx }dx$

Using $\int_0^{π/2} f(x)dx = \int_0^{π/2} f(π/2-x)dx$

This integral is same as $\int_0^{π/2} \frac{sin^2 x +sinx cosx}{1+2sinx cosx}dx$

$\implies{2I= \int_0^{π/2} \frac{1+2sinx cosx}{1+2sinx cosx}dx}$

Answer is $\boxed{I= \frac{π}{4}}$