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

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

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integral of 2cos(ln(x))

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

x sin (ln (x)) + x cos (ln (x)) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \int cos (u) e^{u} d u

Apply Integration By Parts

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

= 2 (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

= 2 (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

= 2 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

Therefore

2 \cdot \int e^{u} cos (u) d u = 2 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

Isolate

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

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

Substitute back

u = ln (x)

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

Simplify

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

= x sin (ln (x)) + x cos (ln (x))

Add a constant to the solution

= x sin (ln (x)) + x cos (ln (x)) + C

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\frac{\partial}{\partial x} (- 5 x y + 2)

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