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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cos(ln(x)))

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

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

Solution steps

\int cos (ln (x)) d x

Apply u-substitution

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

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

Apply Integration By Parts

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

= 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

= 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

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

Therefore

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

Isolate

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

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

Substitute back

u = ln (x)

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

Simplify

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

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

Add a constant to the solution

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

\int sin (2 x) cos^{3} (x) d x

\int 4 sin^{4} (x) cos (x) d x

\int \frac{csc ^{2} ( x )}{1 + 9 cot ^{2} ( x )} d x

\int \frac{9 e ^{9 x}}{x} d x

\int \frac{x arctan ( x ^{2} )}{1 + x ^{4}} d x

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