Question: "I ran across this integration problem that has..."

Aldo Fernandez

Answered question

2022-04-08

I ran across this integration problem that has an interesting pattern.

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

Kendall Wilkinson

Beginner2022-04-09Added 17 answers

If n is even,
then

\frac{1}{\tan^{n} x + 1} = \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x}

is periodic with period

π

Also for

x \in [0, \frac{π}{2}]

we have that for

t = x + \frac{π}{2}

that

\frac{\cos^{n} t}{\sin^{n} t + \cos^{n} t} = \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x}

And so

\int_{0}^{π} \frac{1}{\tan^{n} x + 1} = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} + \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} = \frac{π}{2}

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