What is the integral of cos(ln(mx)) ?

The integral of cos(ln(mx)) is 1/2 x(sin(ln(mx))+cos(ln(mx)))+C

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integral of cos(ln(mx))

\int cos (ln (m x)) d x

\frac{1}{2} x (sin (ln (m x)) + cos (ln (m x))) + C

Solution steps

\int cos (ln (m x)) d x

Apply u-substitution

: \int \frac{cos ( ln ( u ) )}{m} d u

= \int \frac{cos ( ln ( u ) )}{m} d u

Take the constant out:

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

= \frac{1}{m} \cdot \int cos (ln (u)) d u

Apply u-substitution

: \int cos (v) e^{v} d v

= \frac{1}{m} \cdot \int cos (v) e^{v} d v

Apply Integration By Parts

: e^{v} sin (v) - \int e^{v} sin (v) d v

= \frac{1}{m} (e^{v} sin (v) - \int e^{v} sin (v) d v)

Apply Integration By Parts

: - e^{v} cos (v) - \int - e^{v} cos (v) d v

= \frac{1}{m} (e^{v} sin (v) - (- e^{v} cos (v) - \int - e^{v} cos (v) d v))

Take the constant out:

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

= \frac{1}{m} (e^{v} sin (v) - (- e^{v} cos (v) - (- \int e^{v} cos (v) d v)))

Therefore

\frac{1}{m} \cdot \int e^{v} cos (v) d v = \frac{1}{m} (e^{v} sin (v) - (- e^{v} cos (v) - (- \int e^{v} cos (v) d v)))

Isolate

\int e^{v} cos (v) d v

= \frac{1}{m} (\frac{e ^{v} sin ( v )}{2} + \frac{e ^{v} cos ( v )}{2})

Substitute back

= \frac{1}{m} (\frac{e ^{l n (m x)} sin ( ln ( m x ) )}{2} + \frac{e ^{l n (m x)} cos ( ln ( m x ) )}{2})

Simplify

\frac{1}{m} (\frac{e ^{l n (m x)} sin ( ln ( m x ) )}{2} + \frac{e ^{l n (m x)} cos ( ln ( m x ) )}{2}) : \frac{1}{2} x (sin (ln (m x)) + cos (ln (m x)))

= \frac{1}{2} x (sin (ln (m x)) + cos (ln (m x)))

Add a constant to the solution

= \frac{1}{2} x (sin (ln (m x)) + cos (ln (m x))) + C

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What is the integral of cos(ln(mx)) ?
The integral of cos(ln(mx)) is 1/2 x(sin(ln(mx))+cos(ln(mx)))+C