Primitive of x by Tangent of a x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int x \tan a x \rd x = \frac 1 {a^2} \paren {\frac {\paren {a x} ^ 3} 3 + \frac {\paren {a x}^5} {15} + \frac {2 \paren {a x}^7} {105} + \cdots + \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \paren {a x}^{2 n + 1} } {\paren {2 n + 1}!} + \cdots} + C$

where $B_{2 n}$ denotes the $2 n$th Bernoulli number.


Proof

From Power Series Expansion for Tangent Function:


The tangent function has a Taylor series expansion:

\(\ds \tan x\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}\)
\(\ds \) \(=\) \(\ds x + \frac {x^3} 3 + \frac {2 x^5} {15} + \frac {17 x^7} {315} + \cdots\)


where $B_{2 n}$ denotes the Bernoulli numbers.

This converges for $\size x < \dfrac \pi 2$.


\(\ds x \tan ax\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} a^{2 n - 1} x^{2 n} } {\paren {2 n}!}\)
\(\ds \leadsto \ \ \) \(\ds \int x \tan a x \rd x\) \(=\) \(\ds \int \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} a^{2 n - 1} x^{2 n} } {\paren {2 n}!} \rd x\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\int \frac {\paren{-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} a^{2 n - 1} x^{2 n} } {\paren {2 n}!} \rd x}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} a^{2 n - 1} } {\paren {2 n}!} \times \int x^{2 n} \rd x}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} a^{2 n - 1} } {\paren {2 n}!} \times \frac {x^{2 n + 1} } {2 n + 1} + C}\) Primitive of Power
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} a^{2 n - 1} x^{2 n + 1} } {\paren {2 n + 1}!} + C\)
\(\ds \leadsto \ \ \) \(\ds \int x \tan a x \rd x\) \(=\) \(\ds \frac 1 {a ^ 2} \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \paren {a x}^{2 n + 1} } {\paren {2 n + 1}!} + C\)

$\blacksquare$


Also see


Sources