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

Feb 21, 2015

Hello,

Answer. $\frac{3}{2} x \left(\sin \left(\ln \left(x\right)\right) - \cos \left(\ln \left(x\right)\right)\right) + c$, where $c \in \mathbb{R}$.

Use complex numbers.

$\int 3 \sin \left(\ln \left(x\right)\right) \mathrm{dx} = 3 \setminus m a t h \mathfrak{a} k \left\{I m\right\} \setminus \int {e}^{i \ln \left(x\right)} \mathrm{dx} = 3 \setminus m a t h \mathfrak{a} k \left\{I m\right\} \setminus \int {x}^{i} \mathrm{dx}$.

But $\int {x}^{i} \mathrm{dx} = {x}^{i + 1} / \left(i + 1\right) + c = x \frac{{e}^{i \ln \left(x\right)} \left(1 - i\right)}{\left(1 + i\right) \left(1 - i\right)} + c = \frac{x}{2} \left({e}^{i \ln \left(x\right)} \left(1 - i\right)\right)$

Now, you extract the imaginary part of ${e}^{i \ln \left(x\right)} \left(1 - i\right)$ :

${e}^{i \ln \left(x\right)} \left(1 - i\right) = \left(\cos \left(\ln \left(x\right)\right) + i \sin \left(\ln \left(x\right)\right)\right) \left(1 - i\right)$

${e}^{i \ln \left(x\right)} \left(1 - i\right) = i \left(- \cos \left(\ln \left(x\right)\right) + \sin \left(\ln \left(x\right)\right)\right) + \setminus \textrm{s o m e t h \in g r e a l}$

Therefore, $m a t h \mathfrak{a} k \left\{I m\right\} \left({e}^{i \ln \left(x\right)} \left(1 - i\right)\right) = - \cos \left(\ln \left(x\right)\right) + \sin \left(\ln \left(x\right)\right)$.

Conclusion.

$\int 3 \sin \left(\ln \left(x\right)\right) \mathrm{dx} = 3 \frac{x}{2} \left(- \cos \left(\ln \left(x\right)\right) + \sin \left(\ln \left(x\right)\right)\right)$.