MHB How to Solve a Challenging Integration Problem

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

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Congratulations to the following members for their correct solutions::)

1. MarkFL
2. greg1313
3. lfdahl

Solution from MarkFL:

Spoiler

Let:

$$I= \int_{0}^{\frac{\pi}{2}} \frac{\cos^4(x)+\sin(x)\cos^3(x)+\sin^2(x)\cos^2(x)+\sin^3(x)\cos(x)}{\sin^4(x)+\cos^4(x)+2\sin(x)\cos^3 (x)+2\sin^2(x)\cos^2(x)+2\sin^3(x)\cos(x)}\,dx\tag{1}$$

Using the identity:

$$\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$$

Along with co-function identities for sine and cosine and a bit of rearranging, we find:

$$I= \int_{0}^{\frac{\pi}{2}} \frac{\sin^4(x)+\cos(x)\sin^3(x)+\cos^2(x)\sin^2(x)+\cos^3(x)\sin(x)}{\sin^4(x)+\cos^4(x)+2\sin(x)\cos^3 (x)+2\sin^2(x)\cos^2(x)+2\sin^3(x)\cos(x)}\,dx\tag{2}$$

Adding (1) and (2), there results:

$$2I= \int_{0}^{\frac{\pi}{2}} \frac{\sin^4(x)+\cos^4(x)+2\sin(x)\cos^3 (x)+2\sin^2(x)\cos^2(x)+2\sin^3(x)\cos(x)}{\sin^4(x)+\cos^4(x)+2\sin(x)\cos^3 (x)+2\sin^2(x)\cos^2(x)+2\sin^3(x)\cos(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}$$

Hence:

$$I=\frac{\pi}{4}$$

And so in conclusion, we have found:

$$\int_{0}^{\frac{\pi}{2}} \frac{\cos^4(x)+\sin(x)\cos^3(x)+\sin^2(x)\cos^2(x)+\sin^3(x)\cos(x)}{\sin^4(x)+\cos^4(x)+2\sin(x)\cos^3 (x)+2\sin^2(x)\cos^2(x)+2\sin^3(x)\cos(x)}\,dx=\frac{\pi}{4}$$

Solution from greg1313:

Spoiler

$$\begin{align*}I&=\int_0^{\frac{\pi}{2}}\frac{\cos^4\!x+\sin\!x\cos^3\!x+\sin^2\!x\cos^2\!x+\sin^3\!x\cos\!x}{\sin^4\!x+\cos^4\!x+2\sin\!x\cos^3\!x+2\sin^2\!x\cos^2\!x+2\sin^3\!x\cos\!x}\,dx \\
&=\int_0^{\frac{\pi}{2}}\frac{1+\tan\!x+\tan^2\!x+\tan^3\!x}{\tan^4\!x+1+2\tan\!x+2\tan^2\!x+2\tan^3\!x}\,dx \\
&=\int_0^{\frac{\pi}{2}}\frac{\frac{\tan^4\!x-1}{\tan\!x-1}}{\tan\!x\left(\frac{\tan^4\!x-1}{\tan\!x-1}\right)+\frac{\tan^4\!x-1}{\tan\!x-1}}\,dx \\
&=\int_0^{\frac{\pi}{2}}\frac{\cos\!x}{\sin\!x+\cos\!x}\,dx\end{align*}$$

$$\frac{d}{dx}\left(\frac{\cos\!x}{\sin\!x+\cos\!x}\right)=-\frac{1}{(\sin\!x+\cos\!x)^2}$$

Observe that

$$\frac{\cos\!x}{\sin\!x+\cos\!x}$$

has rotational symmetry about the point $$\left(\frac{\pi}{4},\frac12\right)$$ over $$\left[0,\frac{\pi}{2}\right]$$

so we have

$$I=\int_0^{\frac{\pi}{2}}-\frac{2}{\pi}x+1\,dx=\left.\left(-\frac{1}{\pi}x^2+x\right)\right|_0^{\frac{\pi}{2}}=\frac{\pi}{4}$$