What is the integral of 1/t sin(ln(c/t)) ?

The integral of 1/t sin(ln(c/t)) is cos(ln(c/t))+C

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integral of 1/t sin(ln(c/t))

\int \frac{1}{t} sin (ln (\frac{c}{t})) d t

cos (ln (\frac{c}{t})) + C

Solution steps

\int \frac{1}{t} sin (ln (\frac{c}{t})) d t

Apply u-substitution

: \int - sin (u) d u

= \int - sin (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= - (- cos (u))

Substitute back

u = ln (\frac{c}{t})

= - (- cos (ln (\frac{c}{t})))

Simplify

= cos (ln (\frac{c}{t}))

Add a constant to the solution

= cos (ln (\frac{c}{t})) + C

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What is the integral of 1/t sin(ln(c/t)) ?
The integral of 1/t sin(ln(c/t)) is cos(ln(c/t))+C