Expert Answer to "Calculation of ∫0π4(tan⁡(x))ndx"

hionormf

Answered question

2022-01-16

Calculation of

\int_{0}^{\frac{π}{4}} {(\tan (x))}^{n} d x

mauricio0815sh

Beginner2022-01-17Added 34 answers

Evaluate the integral with the recursive relationship below

I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} (x) d x = \int_{0}^{\frac{π}{4}} \tan^{n - 2} x (\sec^{2} x - 1) d x

= \int_{0}^{\frac{π}{4}} \tan^{n - 2} x d (\tan x) - \int_{0}^{\frac{π}{4}} \tan^{n - 2} x d x

= \frac{1}{n - 1} \tan^{n - 1} x ∣_{0}^{\frac{π}{4}} - I_{n - 2} = \frac{1}{n - 1} - I_{n - 2}

with

I_{0} = \frac{π}{4}

and

I_{1} = \frac{12}{\ln 2}

Wendy Boykin

Beginner2022-01-18Added 35 answers

\int_{0}^{\frac{π}{4}} {(\tan (x))}^{n} d x \overset{x \to \arctan z}{=} \int_{0}^{1} \frac{z^{n}}{z^{2} + 1} d z = \int_{0}^{1} (z^{n} - z^{n + 2} + z^{n + 4} - \dots) d z

equals

\frac{1}{n + 1} - \frac{1}{n + 3} + \frac{1}{n + 5} - \dots

i.e. a tail of

\sum_{m \geq 0} \frac{{(- 1)}^{m}}{2 m + 1} = \frac{π}{4}

\sum_{m \geq 0} \frac{{(- 1)}^{m}}{2 m + 2} = \log \sqrt{2}

according to the parity of n.

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