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

y = 3 x, x = 4, y = \frac{3}{x}

\int \frac{5}{9 - 16 x ^{2}} d x

\frac{d y}{d t} = - k y

x \to - 1 lim (- x^{2} + 4 x - 5)

\frac{d}{d x} (\frac{2 x ^{2}}{( x ^{2} + x ) ^{3}})

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