What is the integral of xcos(ln(x)) ?

The integral of xcos(ln(x)) is 1/5 x^2sin(ln(x))+2/5 x^2cos(ln(x))+C

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integral of xcos(ln(x))

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

\frac{1}{5} x^{2} sin (ln (x)) + \frac{2}{5} x^{2} cos (ln (x)) + C

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= \frac{e ^{2 u} sin ( u )}{5} + \frac{2 e ^{2 u} cos ( u )}{5}

Substitute back

u = ln (x)

= \frac{e ^{2 l n (x)} sin ( ln ( x ) )}{5} + \frac{2 e ^{2 l n (x)} cos ( ln ( x ) )}{5}

Simplify

\frac{e ^{2 l n (x)} sin ( ln ( x ) )}{5} + \frac{2 e ^{2 l n (x)} cos ( ln ( x ) )}{5} : \frac{1}{5} x^{2} sin (ln (x)) + \frac{2}{5} x^{2} cos (ln (x))

= \frac{1}{5} x^{2} sin (ln (x)) + \frac{2}{5} x^{2} cos (ln (x))

Add a constant to the solution

= \frac{1}{5} x^{2} sin (ln (x)) + \frac{2}{5} x^{2} cos (ln (x)) + C

\int x^{14} e^{x^{3}} d x

\frac{\partial}{\partial x} (e^{x l n (y)})

x \to 2 lim (\frac{x}{( x - 2 ) ^{2}})

2 x^{4} - 11 x^{2}

\frac{\partial}{\partial x} (\frac{G ( x , y , r ) x y}{r ^{2}})

What is the integral of xcos(ln(x)) ?
The integral of xcos(ln(x)) is 1/5 x^2sin(ln(x))+2/5 x^2cos(ln(x))+C