MHB Can Trigonometric Identities Simplify This Week's Complex Integral?

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The discussion centers on evaluating the integral \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. Participants are encouraged to apply trigonometric identities to simplify the integral. Members castor28 and lfdahl are recognized for providing correct solutions. The thread emphasizes the importance of following the Problem of the Week guidelines for submissions. The discussion highlights the role of trigonometric identities in solving complex integrals effectively.
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Here is this week's POTW:

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

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

1. castor28
2. lfdahl

Solution from castor28:
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 $I = \dfrac\pi4$.
 
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