Tuesday, June 2, 2015

Putnam Contest Problem: Evaluate [MATH]\tiny \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].

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 provided by Mark, another contributor of this blog.

Let:

[MATH]\small 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-(1)[/MATH]

Using the identity:

[MATH]\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx[/MATH]

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

[MATH]\small 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-(2)[/MATH]

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

[MATH]\small 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}[/MATH]

Hence:

[MATH]I=\frac{\pi}{4}[/MATH]

And so in conclusion, we have found:

[MATH]\small \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}[/MATH]