Evaluate ∫sin(logx)dx
ex2[sin(logx)−cos(logx)]+c
x2[cos(logx)−sin(logx)]+c
x2[sin(logx)−cos(logx)]+c
None of these
Let I=∫sin (logx)dx Let logx=t . Then x=et or dx=etdt . Thus. I=∫sint etdt=et2(sint−cost)+c using the formula ∫ eaxsin(bx+c)=eaxa2+b2[asin(bx+c)-bcos(bx+c)] Hence, ∫sin(logx)dx=x2[sin(logx)−cos(logx)]+c