Can Trigonometric Identities Simplify This Week's Complex Integral?

[anemone]

Here is this week's POTW:

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Evaluate \(\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\cos^4x+\sin x\cos^3 x+\sin^2 x \cos^2 x+\sin^3 x\cos x}{\sin^4 x+\cos^4 x+2\sin x\cos^3 x+2\sin^2 x \cos^2 x+2\sin^3 x\cos x}\,dx\).

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. castor28
2. lfdahl

Solution from castor28:

Spoiler

Let us write $I$ for the integral to be evaluated. The integrand can be written as:
$$
1 - \frac{\sin^4x+\sin x\cos^3 x+\sin^2 x \cos^2 x+\sin^3 x\cos x}{\sin^4 x+\cos^4 x+2\sin x\cos^3 x+2\sin^2 x \cos^2 x+2\sin^3 x\cos x}
$$
Using the substitution $x\to\frac{\pi}{2}-x$, the integral of the second term becomes:
\begin{align*}
&-\int_{\frac\pi2}^0\frac{\cos^4x+\cos x\sin^3 x+\cos^2 x \sin^2 x+\cos^3 x\sin x}{\cos^4 x+\sin^4 x+2\cos x\sin^3 x+2\cos^2 x \sin^2 x+2\cos^3 x\sin x}\,dx\\
&= +\int_0^{\frac\pi2}\frac{\cos^4x+\cos x\sin^3 x+\cos^2 x \sin^2 x+\cos^3 x\sin x}{\cos^4 x+\sin^4 x+2\cos x\sin^3 x+2\cos^2 x \sin^2 x+2\cos^3 x\sin x}\,dx\\
&= I
\end{align*}
This leaves us with:
$$
I = \int_0^{\frac\pi2}dx - I
$$
which gives $I = \dfrac\pi4$.