What is the integral of-12cos(ln(x)) ?

The integral of-12cos(ln(x)) is -12(1/2 xsin(ln(x))+1/2 xcos(ln(x)))+C

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integral of-12cos(ln(x))

\int - 12 cos (ln (x)) d x

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

Solution steps

\int - 12 cos (ln (x)) d x

Take the constant out:

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

= - 12 \cdot \int cos (ln (x)) d x

Apply u-substitution

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

= - 12 \cdot \int cos (u) e^{u} d u

Apply Integration By Parts

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

= - 12 (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

= - 12 (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

= - 12 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

Therefore

- 12 \cdot \int e^{u} cos (u) d u = - 12 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

Isolate

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

= - 12 (\frac{e ^{u} sin ( u )}{2} + \frac{e ^{u} cos ( u )}{2})

Substitute back

u = ln (x)

= - 12 (\frac{e ^{l n (x)} sin ( ln ( x ) )}{2} + \frac{e ^{l n (x)} cos ( ln ( x ) )}{2})

Simplify

- 12 (\frac{e ^{l n (x)} sin ( ln ( x ) )}{2} + \frac{e ^{l n (x)} cos ( ln ( x ) )}{2}) : - 12 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))

= - 12 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))

Add a constant to the solution

= - 12 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))) + C

\frac{\partial}{\partial t} (1 + (t - y)^{2})

h \to 0 lim (\frac{cos ( 6 h ) - 1}{h})

\int \frac{x ^{3} + 4 x ^{2} + x - 1}{x ^{3} + x ^{2}} d x

y^{^{''}} + y = 2 + x^{2}

\frac{d ^{2} y}{d x ^{2}} = 49 y

What is the integral of-12cos(ln(x)) ?
The integral of-12cos(ln(x)) is -12(1/2 xsin(ln(x))+1/2 xcos(ln(x)))+C