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Question Number 151486 by mathocean1 last updated on 21/Aug/21
∫ln(sinx)=?x≠0+2kπ;k∈Z
Answered by Olaf_Thorendsen last updated on 21/Aug/21
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
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