How do you integrate $\int 3 sin (ln (x)) d x$ $∫3sin(ln(x))dx$ ?

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Answer 1 · 2015-02-21T02:21:24

Hello,

Answer. $\frac{3}{2} x (sin (ln (x)) - cos (ln (x))) + c$ $3/2 x (sin(ln(x)) -cos(ln(x))) + c$ , 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.

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

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