## Saturday, April 25, 2015

### Putnam Definite Integral Hard Problem: Evaluate $\displaystyle\int_0^{\frac{\pi}{2}}\frac{dx}{1+(\tan x)^{\sqrt{2}}}$.

Another hard definite integral but it is not as bad as it looks. Let's first try the substitution method:

Attempt I:

We are given to evaluate:

[MATH]I=\int_0^{\pi/2}\frac{1}{1+\left(\tan(x)\right)^{\sqrt{2}}}\,dx[/MATH]

If we use the substitution:

[MATH]u=\frac{\pi}{2}-x[/MATH]

and then use $x$ as the dummy variable instead, we obtain:

[MATH]I=\int_0^{\pi/2}\frac{\left(\tan(x)\right)^{\sqrt{2}}}{1+\left(\tan(x)\right)^{\sqrt{2}}}\,dx[/MATH]

And so adding the two expressions, we obtain:

[MATH]2I=\int_0^{\pi/2}1\,dx=\left[x\right]_0^{\pi/2}=\frac{\pi}{2}[/MATH]

Hence:

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

Attempt II:

On another attempt, we would apply one of the properties of definite integrals in which we have

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

We are given to evaluate:

[MATH]I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^\sqrt{2}x }\,dx[/MATH]

Using the property stated above and a co-function identity $\tan \left(\frac{\pi}{2}-x\right)=\cot x$, we may state:

[MATH]I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^\sqrt{2}(x)}\,dx[/MATH]

Adding the two equations, we obtain:

[MATH]2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^\sqrt{2} x }+\frac{1}{1+\cot^\sqrt{2} x}\,dx[/MATH]

[MATH]2I=\int_0^{\frac{\pi}{2}}\frac{\cancel{2+\tan^\sqrt{2} x+\cot^\sqrt{2} x}}{\cancel{2 + \tan^\sqrt{2} x+\cot^\sqrt{2} x}}\,dx=\int_0^{\frac{\pi}{2}}1\,dx=\frac{\pi}{2}[/MATH]

Hence:

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