Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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......

Terms of Service

Privacy Policy

Contact: info@tinkutara.com