I need to calculate $\int_{0}^{\pi\over 2}{dx\over {1+\sqrt{\tan x}}}$

In this section of the course where this problem is, we have seen 2 techniques: integration by parts and substitution, so I am assuming the solution has to make use of one of those somewhere.

  • Attempt to use integration by parts: We define $f(x):={1\over 1+\sqrt{\tan x}}$ and $g(x):=x$

We then have $\int_{0}^{\pi\over 2}{dx\over {1+\sqrt{\tan x}}}=\int_{0}^{\pi\over 2}fg'=(fg)({\pi\over 2})-(fg)(0)-\int_{0}^{\pi\over 2}f'g$

but f'(x) is a relatively complicated expression and doing this seems to only make matters worse.

  • Attempt to use substitution: We define $g(x):=\sqrt{\tan x}$ in an attempt to get the form $\int_{0}^{\pi\over 2}{g'(x)\over 1+g(x)}$

but then again this derivative is a complicated expression which only seems to make the problem tougher. I am just not seeing how to simplify.

  • $\begingroup$ Try the substitution $t=\pi/2-x$ $\endgroup$
    – Tom Himler
    Oct 29, 2019 at 16:18
  • $\begingroup$ The result should be $$\frac{\pi}{4}$$ $\endgroup$ Oct 29, 2019 at 16:19

1 Answer 1


Let $$I(a)=\int_{0}^\frac{\pi}{2}\frac{1}{1+\tan^a(x)}dx$$ Enforcing the substitution $t=\frac{\pi}{2}-x\implies dt=-dx$ $$I(a)=\int_{0}^\frac{\pi}{2}\frac{1}{1+\cot^a(t)}dt=\int_{0}^\frac{\pi}{2}\frac{\tan^a(t)}{1+\tan^a(t)}dt$$ Adding the two representations of $I(a)$: $$2I(a)=\int_{0}^\frac{\pi}{2}dt=\frac{\pi}{2}$$ Solving for $I(a)$: $$I(a)=\frac{\pi}{4}$$ Plugging $a=\frac{1}{2}\implies I\left(\frac{1}{2}\right)=\frac{\pi}{4}$

  • $\begingroup$ I am pretty sure this problem appeared on the Putnam before. $\endgroup$
    – Bananas
    Oct 31, 2019 at 9:53

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