What is the integral of-11cos(ln(x)) ?

The integral of-11cos(ln(x)) is -11(1/2 xsin(ln(x))+1/2 xcos(ln(x)))+C

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integral of-11cos(ln(x))

\int - 11 cos (ln (x)) d x

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

Solution steps

\int - 11 cos (ln (x)) d x

Take the constant out:

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

= - 11 \cdot \int cos (ln (x)) d x

Apply u-substitution

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

= - 11 \cdot \int cos (u) e^{u} d u

Apply Integration By Parts

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

= - 11 (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

= - 11 (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

= - 11 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

Therefore

- 11 \cdot \int e^{u} cos (u) d u = - 11 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

Isolate

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

= - 11 (\frac{e ^{u} sin ( u )}{2} + \frac{e ^{u} cos ( u )}{2})

Substitute back

u = ln (x)

= - 11 (\frac{e ^{l n (x)} sin ( ln ( x ) )}{2} + \frac{e ^{l n (x)} cos ( ln ( x ) )}{2})

Simplify

- 11 (\frac{e ^{l n (x)} sin ( ln ( x ) )}{2} + \frac{e ^{l n (x)} cos ( ln ( x ) )}{2}) : - 11 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))

= - 11 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))

Add a constant to the solution

= - 11 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))) + C

f (x) = \frac{x}{x + 1}, at x = 0

f (x) = (2 x - 1) (6 x - 6)^{- 1}

f (x) = \frac{x ^{2}}{x + 2}

(tan (y)) d x - (cot (x)) d y = 0

\int \frac{x - 9}{( x + 5 ) ( x - 2 )} d x

What is the integral of-11cos(ln(x)) ?
The integral of-11cos(ln(x)) is -11(1/2 xsin(ln(x))+1/2 xcos(ln(x)))+C