What is the integral of cos(ln(2x)) ?

The integral of cos(ln(2x)) is 1/2 (xsin(ln(2x))+xcos(ln(2x)))+C

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integral of cos(ln(2x))

\int cos (ln (2 x)) d x

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

Solution steps

\int cos (ln (2 x)) d x

Apply u-substitution

= \int cos (ln (u)) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (ln (u)) d u

Apply u-substitution

= \frac{1}{2} \cdot \int cos (v) e^{v} d v

Apply Integration By Parts

= \frac{1}{2} (e^{v} sin (v) - \int e^{v} sin (v) d v)

Apply Integration By Parts

= \frac{1}{2} (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}{2} (e^{v} sin (v) - (- e^{v} cos (v) - (- \int e^{v} cos (v) d v)))

Therefore

\frac{1}{2} \cdot \int e^{v} cos (v) d v = \frac{1}{2} (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}{2} (\frac{e ^{v} sin ( v )}{2} + \frac{e ^{v} cos ( v )}{2})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x}{x ^{2} - 2 x + 5} d x

\frac{d}{d x} (arccot (\frac{x}{4}))

\int \frac{y ^{3}}{( 2 + 3 y ^{4} ) ^{3}} d y

sin (tan (3 x))

lo g_{4} (e^{2 x} cos (3 x))

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