Josalynn

2021-10-15

Evaluate the following integrals. Include absolute values only when needed.
$\int \frac{\mathrm{sin}\left(\mathrm{ln}x\right)}{4x}dx$

Talisha

It is given that, $\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx$
We have to evaluate it.
We have, $\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx$
Let $u=\mathrm{ln}\left(x\right)$
differentiate equation w.r.t x we get
$du=\frac{1}{x}dx,⇒dx=xdu$
Then equation becomes
$⇒\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx=\frac{1}{4}\int \mathrm{sin}\left(u\right)du$
$⇒\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx=\frac{1}{4}\left(-\mathrm{cos}\left(u\right)\right)+C$, where C is arbitrary constant
Putting the value of $u=\mathrm{ln}\left(x\right)$ in above equation, we get
$⇒\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx=\frac{1}{4}\left(-\mathrm{cos}\left(u\right)\right)+C$
Hence, $\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx=\frac{\mathrm{cos}\left(\mathrm{ln}\left(x\right)\right)}{4}+C$

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