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

The integral of cos(ln(y)) is 1/2 ysin(ln(y))+1/2 ycos(ln(y))+C

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integral of cos(ln(y))

\int cos (ln (y)) d y

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

Solution steps

\int cos (ln (y)) d y

Apply u-substitution

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

= \int cos (u) e^{u} d u

Apply Integration By Parts

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

= e^{u} sin (u) - \int e^{u} sin (u) d u

Apply Integration By Parts

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

= e^{u} sin (u) - (- e^{u} cos (u) - \int - e^{u} cos (u) d u)

Take the constant out:

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

= e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u))

Therefore

\int e^{u} cos (u) d u = e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u))

Isolate

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

= \frac{e ^{u} sin ( u )}{2} + \frac{e ^{u} cos ( u )}{2}

Substitute back

u = ln (y)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of cos(ln(y)) ?
The integral of cos(ln(y)) is 1/2 ysin(ln(y))+1/2 ycos(ln(y))+C