How to Evaluate this integral and this is a practice problem of Bangladesh math Olympiad team selection : First I tried to simplify it but confused , where to start :
$$\int^{\pi/2}_0 \frac{\cos^4x + \sin x \cos^3 x + \sin^2x\cos^2x + \sin^3x\cos x}{\sin^4x + \cos^4x + 2\sin x\cos^3x + 2\sin^2x\cos^2x + 2\sin^3x\cos x} dx$$
Hint:
Use the reflection property. Replace $x\mapsto \pi/2-x$. Adding the two expressions $2I=\pi/2$. And $I=\pi/4$.
Hint. To simplify the integrand, write $c=\cos x$ and $s=\sin x$ and notice that the denominator is of the form $$c^4+s^4+2cs(c^2+s^2+cs),$$ the numerator also being of this form except that the third term has coefficient $1$ in the place of the $2.$ Thus we simplify both simultaneously by focusing on the denominator, which becomes (since $c^2+s^2=1$) $$c^4+s^4+2cs(1+cs)=(c^2)^2+2c^2s^2+(s^2)^2+2cs=(c^2+s^2)^2+2cs=1+2cs=1+\sin 2x.$$
The numerator is $$c^4+cs(c^2+s^2+cs)=c^4+cs(1+cs)=(c^2+s^2)^2-2(cs)^2+cs+(cs)^2-s^4==1+cs-(cs)^2-s^4=1+cs-s^2(c^2+s^2)=1-s^2+cs=c^2+cs=c(c+s)=\cos x(\cos x+\sin x),$$ from where you should now be able to proceed.
The problem is just largely algebra.
Simplify the top and the bottom:
$$\text{top}= \cos x \cdot (\cos x + \sin x)$$
$$\text{bottom}= (\cos{x} + \sin{x})^2$$
Thus $$ I = \int_0^{\pi/2} \frac{\cos{x}}{\cos{x}+\sin{x}} \,\mathrm{dx} $$
Which by symmetry is equal to
$$ I = \int_0^{\pi/2} \frac{\sin{x}}{\cos{x}+\sin{x}} \,\mathrm{dx} $$
So that $$2I = \int_0^{\pi/2} \frac{\sin{x}+\cos{x}}{\cos{x}+\sin{x}} \,\mathrm{dx} = \int_0^{\pi/2} \mathrm{dx} = \frac{\pi}{2}. $$
So that $$I = \frac{\pi}{4}.$$
