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I would like to find the closed form of:

$$\int_0^{\pi/2} \frac 1 {1+\tan^{s} x} \, dx$$

I have tried substitution method, $t=\tan (\frac{x}{2})$ Then I ended up with with a very complicated rational function in the integrand.

Can anyone help?

Guy Fsone
  • 23,903

1 Answers1

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Using the change of variable: $u=\frac\pi2-x $ and since, $\tan x =\cot(\frac\pi2 -x)$ we have, \begin{align} & \int_0^{\frac\pi2}\frac{1}{1+\tan^{s} x} \, dx = \int_0^{\frac\pi2}\frac{1}{1+\tan^{s} (\frac\pi2-u) } \, du \\[10pt] = {} & \int_0^{\frac\pi2}\frac{1}{1+\cot^{s}u} \, du = \int_0^{\frac\pi2}\frac{\tan^{s} u}{1+\tan^{s} u} \, du \color{red}{= \frac{\pi}{2} -\int_0^{\frac\pi2}\frac{1}{1+\tan^{s} u} \, du} \end{align}

That is $$\int_0^{\frac\pi2}\frac{1}{1+\tan^{s} x} \, dx =\frac\pi4$$