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a
ex2[sin(logx)−cos(logx)]+c
b
x2[cos(logx)−sin(logx)]+c
c
x2[sin(logx)−cos(logx)]+c
d
None of these
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Correct option is C
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)]+cSimilar Questions
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