What is the integral of cos(ln(7x)) ?

The integral of cos(ln(7x)) is 1/2 x(sin(ln(7x))+cos(ln(7x)))+C

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integral of cos(ln(7x))

\int cos (ln (7 x)) d x

\frac{1}{2} x (sin (ln (7 x)) + cos (ln (7 x))) + C

Solution steps

\int cos (ln (7 x)) d x

Apply u-substitution

: \int cos (ln (u)) \frac{1}{7} d u

= \int cos (ln (u)) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int cos (ln (u)) d u

Apply u-substitution

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

= \frac{1}{7} \cdot \int cos (v) e^{v} d v

Apply Integration By Parts

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

= \frac{1}{7} (e^{v} sin (v) - \int e^{v} sin (v) d v)

Apply Integration By Parts

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

= \frac{1}{7} (e^{v} sin (v) - (- e^{v} cos (v) - \int - e^{v} cos (v) d v))

Take the constant out:

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

= \frac{1}{7} (e^{v} sin (v) - (- e^{v} cos (v) - (- \int e^{v} cos (v) d v)))

Therefore

\frac{1}{7} \cdot \int e^{v} cos (v) d v = \frac{1}{7} (e^{v} sin (v) - (- e^{v} cos (v) - (- \int e^{v} cos (v) d v)))

Isolate

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

= \frac{1}{7} (\frac{e ^{v} sin ( v )}{2} + \frac{e ^{v} cos ( v )}{2})

Substitute back

= \frac{1}{7} (\frac{e ^{l n (7 x)} sin ( ln ( 7 x ) )}{2} + \frac{e ^{l n (7 x)} cos ( ln ( 7 x ) )}{2})

Simplify

\frac{1}{7} (\frac{e ^{l n (7 x)} sin ( ln ( 7 x ) )}{2} + \frac{e ^{l n (7 x)} cos ( ln ( 7 x ) )}{2}) : \frac{1}{2} x (sin (ln (7 x)) + cos (ln (7 x)))

= \frac{1}{2} x (sin (ln (7 x)) + cos (ln (7 x)))

Add a constant to the solution

= \frac{1}{2} x (sin (ln (7 x)) + cos (ln (7 x))) + C

x \to 0 lim (- x^{6})

x \to 4 - lim (\frac{x}{x + 4})

\int \frac{x ^{2}}{( x ^{3} - 1 ) ^{2}} d x

x \to 0 lim (x ln (\frac{1}{x}))

\frac{1}{( s - 1 ) ^{2} \cdot ( s + 5 )}

What is the integral of cos(ln(7x)) ?
The integral of cos(ln(7x)) is 1/2 x(sin(ln(7x))+cos(ln(7x)))+C