Menu Close

ln-sinx-x-0-2kpi-k-Z-

Tinku Tara 0 Comments
Pin




Question Number 151486 by mathocean1 last updated on 21/Aug/21
∫ln(sinx)=? x ≠ 0 + 2kπ ; k ∈ Z
∫ln(sinx)=?x≠0+2kπ;k∈Z
Answered by Olaf_Thorendsen last updated on 21/Aug/21
F(x) = ∫ln(sinx) dx F(x) = xln(sinx)−∫xcotx dx F(x) = xln(sinx)−∫xΣ_(n=0) ^∞ (((−1)^n 2^(2n) B_(2n) x^(2n−1) )/((2n)!)) dx F(x) = xln(sinx)−∫Σ_(n=0) ^∞ (((−1)^n 2^(2n) B_(2n) x^(2n) )/((2n)!)) dx F(x) = xln(sinx)−Σ_(n=0) ^∞ (((−1)^n 2^(2n) B_(2n) x^(2n+1) )/((2n+1)!))+C
F(x)=∫ln(sinx)dxF(x)=xln(sinx)−∫xcotxdxF(x)=xln(sinx)−∫x∑∞n=0(−1)n22nB2nx2n−1(2n)!dxF(x)=xln(sinx)−∫∑∞n=0(−1)n22nB2nx2n(2n)!dxF(x)=xln(sinx)−∑∞n=0(−1)n22nB2nx2n+1(2n+1)!+C

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *