Friday, June 12, 2015

Work backwards Putnam IMO Contest Problem

Putnam Contest Problem:

Evaluate [MATH]\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[/MATH].

Solution of mine, which is different from the previously posted method provided by Mark(putnam-contest-problem):

Let [MATH]I=\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[/MATH].

[MATH]\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 \\
&=\frac{(\cos^4\!x+\sin\!x\cos^3\!x+\sin^2\!x\cos^2\!x+\sin^3\!x\cos\!x)\div(\cos^4 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)\div(\cos^4 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{1+\tan\!x+\tan^2\!x+\tan^3\!x}{(\tan^4\!x+\tan^3\!x+\tan^2\!x+\tan\!x)+(\tan^3\!x+\tan^2\!x+\tan\!x+1)}\,dx \\
&=\int_0^{\frac{\pi}{2}}\frac{(1+\tan\!x+\tan^2\!x+\tan^3\!x)}{\tan\!x(\tan^3\!x+\tan^2\!x+\tan\!x+1)+(\tan^3\!x+\tan^2\!x+\tan\!x+1)}\,dx \\
&=\int_0^{\frac{\pi}{2}}\frac{1}{\tan\!x+1}\,dx \\
&=\int_0^{\frac{\pi}{2}}\frac{\cos\!x}{\sin\!x+\cos\!x}\,dx\end{align*}[/MATH]

My solution is based on work backwards method, because I noticed whenever the integrand is of rational function with the denominator function as the sum of sine and cosine function, then I know the its anti-derivative must contain the natural logarithm function of the sum of sine and cosine function as well, that is to say, I realized:

$\dfrac{d}{dx}\ln (\sin x+\cos x)=\dfrac{\cos x-\sin x}{\sin x+\cos x}$;$\dfrac{d}{dx}(x)=1$

And

\begin{align*}1+\dfrac{\cos x-\sin x}{\sin x+\cos x}&=\dfrac{\sin x+\cos x+\cos x-\sin x}{\sin x+\cos x}\\&=\dfrac{2\cos x}{\sin x+\cos x}\end{align*}

Therefore, it's not hard to combine them both wisely to get

\begin{align*}\dfrac{d}{dx}\left(\dfrac{x}{2}+\ln (\sin x+\cos x)\right)&=\dfrac{1}{2}+\dfrac{\cos x-\sin x}{\sin x+\cos x}\\&=\dfrac{1}{2}\left(\dfrac{2\cos x}{\sin x+\cos x}\right)\\&=\dfrac{\cos x}{\sin x+\cos x}\end{align*}

Hence, we have:

[MATH]\begin{align*}I&=\int_0^{\frac{\pi}{2}}\frac{\cos\!x}{\sin\!x+\cos\!x}\,dx\\&=\left(\dfrac{x}{2}+\ln (\sin x+\cos x)\right)_0^{\frac{\pi}{2}}\\&=\left(\dfrac{\pi}{4}+\ln (1+0)\right)-\left(0+\ln (0+1)\right)\\&=\dfrac{\pi}{4}\end{align*}[/MATH]