What is the integral of cos(ln(x^2)) ?

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

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integral of cos(ln(x^2))

\int cos (ln (x^{2})) d x

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

Solution steps

\int cos (ln (x^{2})) d x

Simplify

ln (x^{2}) : 2 ln (x)

= \int cos (2 ln (x)) d x

Apply u-substitution

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

Apply Integration By Parts

= e^{u} \frac{1}{2} sin (2 u) - \int e^{u} \frac{1}{2} sin (2 u) d u

Take the constant out:

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

= e^{u} \frac{1}{2} sin (2 u) - \frac{1}{2} \cdot \int e^{u} sin (2 u) d u

Apply Integration By Parts

= e^{u} \frac{1}{2} sin (2 u) - \frac{1}{2} (- \frac{1}{2} e^{u} cos (2 u) - \int - \frac{1}{2} e^{u} cos (2 u) d u)

Take the constant out:

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

= e^{u} \frac{1}{2} sin (2 u) - \frac{1}{2} (- \frac{1}{2} e^{u} cos (2 u) - (- \frac{1}{2} \cdot \int e^{u} cos (2 u) d u))

Therefore

\int e^{u} cos (2 u) d u = e^{u} \frac{1}{2} sin (2 u) - \frac{1}{2} (- \frac{1}{2} e^{u} cos (2 u) - (- \frac{1}{2} \cdot \int e^{u} cos (2 u) d u))

Isolate

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

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

Substitute back

u = ln (x)

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

Simplify

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

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

Add a constant to the solution

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

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