How do you integrate #∫3sin(ln(x))dx#?

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Answer 1 · 2015-02-21T02:21:24

Hello,

Answer. #3/2 x (sin(ln(x)) -cos(ln(x))) + c#, where #c in RR#.

Use complex numbers.

#int 3 sin(ln(x)) dx = 3 \mathfrak{Im} \int e^{i ln (x)} dx = 3 \mathfrak{Im}\int x^(i) dx#.

But #int x^(i) dx = x^(i+1)/(i+1) + c = x (e^(i ln(x))(1-i))/((1+i)(1-i)) + c = x/2 (e^(i ln(x))(1-i))#

Now, you extract the imaginary part of #e^(i ln(x))(1-i)# :

#e^(i ln(x))(1-i) = (cos(ln(x)) + i sin(ln(x)))(1-i)#

#e^(i ln(x))(1-i) = i(-cos(ln(x)) + sin(ln(x))) + \text{something real}#

Therefore, #mathfrak{Im}(e^(i ln(x))(1-i)) = -cos(ln(x)) + sin(ln(x))#.

Conclusion.

#int 3 sin(ln(x)) dx = 3 x/2(-cos(ln(x)) + sin(ln(x)))#.