Question Number 114880 by manuel2456 last updated on 21/Sep/20

∫ln (sin (x))dx=?

Commented byMJS_new last updated on 21/Sep/20

similar to question 113634

Answered by mathdave last updated on 23/Sep/20

solution   from binomial theorem  let   (1/(1−y))=Σ_(n=1) ^∞ y^(n−1)  or ln(1−y)=−Σ_(n=1) ^∞ (y^n /n)  put y=e^(i2x)   ln(1−e^(i2x) )=−Σ_(n=1) ^∞ (e^(i2nx) /n)  ln(1−(cos2x+isin2x))=−Σ_(n=1) ^∞ (1/n)(cos(2nx)+isin(2nx))  but 2sin^2 x=1−cos2x  and sin2x=2sinxcosx  ln(2sin^2 x−i2sinxcosx)=−Σ_(n=1) ^∞ ((cos(2nx))/n)−iΣ_(n=1) ^∞ ((sin(2nx))/n)  ln(2sinx(sinx−icosx))=−Σ_(n=1) ^∞ ((cos(2nx))/n)−iΣ_(n=1) ^∞ ((sin(2nx))/n)  ln(2sinx)+ln(sinx−icosx)=−Σ_(n=1) ^∞ ((cos(2nx))/n)−iΣ_(n=1) ^∞ ((sin(2nx))/n)  let  ln(−i(−(1/i)sinx+cosx))=ln(−i(cosx−(1/i)sinx))=ln(−i(cosx+isinx))=ln(−i.e^(ix) )  but  ln(−i.e^(ix) )=ln(−1)+ln(e^(ix) )=ix+ln(−i)=ix−i(π/2)=−i((π/2)−x)  ∴ln(2sinx)−i((π/2)−x)=−Σ_(n=1) ^∞ ((cos(2nx))/n)−iΣ_(n=1) ^∞ ((sin(2nx))/n)  by equating real and imaginary part  ln(2sinx)=−Σ_(n=1) ^∞ ((cos(2nx))/n)  ln2+ln(sinx)=−Σ_(n=1) ^∞ ((cos(2nx))/n)  ln(sinx)=−ln2−Σ_(n=1) ^∞ ((cos(2nx))/n) ......(1)    ∵∫ln(sinx)dx=−ln2∫dx−Σ_(n=1) ^∞ (1/n)∫cos(2nx)dx  ∵∫ln(sinx)dx=−xln2−Σ_(n=1) ^∞ (1/(2n^2 ))sin(2nx)  by mathdave(23/09/2020)

Commented byTawa11 last updated on 06/Sep/21

grest sir