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ideklaraz7xz

Answered question

2022-03-25

Why does

\int_{0}^{n π} \frac{d x}{1 + \tan^{2 k} (x)} = n \frac{π}{2}

hold for all non-negative integer k?

Ashton Conrad

Beginner2022-03-26Added 11 answers

\int_{0}^{n π} \frac{d x}{1 + \tan^{2 k} x}

The integrand is periodic with period

π

and symmetric within a period about

\frac{π}{2}

= 2 n \int_{0}^{\frac{π}{2}} \frac{d x}{1 + \tan^{2 k} x}

Swap the integral bounds. The integrand then evaluates to

\frac{\tan^{2 k}}{1 + \tan^{2 k} x}

\tan (\frac{π}{2} - x) = \cot x

The sum of the two integrands is 1:

= \frac{2 n}{2} \int_{0}^{\frac{π}{2}} 1, d x = \frac{n π}{2}

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