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:

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

If we use the substitution:

$\displaystyle u=\frac{\pi}{2}-x$

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

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

And so adding the two expressions, we obtain:

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

Hence:

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

Attempt II:

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

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

We are given to evaluate:

$\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^\sqrt{2}x }\,dx$

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

$\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^\sqrt{2}(x)}\,dx$

Adding the two equations, we obtain:

$\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^\sqrt{2} x }+\frac{1}{1+\cot^\sqrt{2} x}\,dx$

$\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{\cancel{2+\tan^\sqrt{2} x+\cot^\sqrt{2} x}}{\cancel{2 + \tan^\sqrt{2} x+\cot^\sqrt{2} x}}\,dx=\int_0^{\frac{\pi}{2}}1\,dx=\frac{\pi}{2}$

Hence:

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