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0-pi-2-x-sin-2-x-ln-sin-x-dx-




Question Number 155252 by mnjuly1970 last updated on 27/Sep/21
     ∫_0 ^( (π/2)) x.sin^( 2) (x).ln(sin(x))dx
0π2x.sin2(x).ln(sin(x))dx
Answered by phanphuoc last updated on 28/Sep/21
lnsinx=−ln2−Σ_(k=1) ^∞ ((cos(2kx))/k)  →I=∫_0 ^∞ xsin^2 x(−ln2−Σ((cos(2kx))/k))dx  =−ln2∫_0 ^∞ xsin^2 xdx−Σ(1/(2k))x(1−cos2x)cos(2kx)  you can IBP→......
lnsinx=ln2k=1cos(2kx)kI=0xsin2x(ln2Σcos(2kx)k)dx=ln20xsin2xdxΣ12kx(1cos2x)cos(2kx)youcanIBP

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