Math, asked by amritapati123, 7 months ago

integration of sin(ln x)dx

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Answered by vaishanavi2003

I=∫sin(logx)×1dx=sin(logx)×x−∫cos(logx)×(x1)×xdx=xsin(logx)−∫cos(logx)×1dx=xsin(logx)−[cos(logx)×x−∫sin(logx)×(x1)×xdx]∴I=xsin(logx)−cos(logx)×x−∫sin(logx)dxor2I=x[sin(logx)−cos(logx)]∴I=(2x)[sin(logx)−cos(logx)]

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