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

The integral of (cos(ln(x))sin(ln(x)))/x is -1/4 cos(2ln(x))+C

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integral of (cos(ln(x))sin(ln(x)))/x

\int \frac{cos ( ln ( x ) ) sin ( ln ( x ) )}{x} d x

- \frac{1}{4} cos (2 ln (x)) + C

Solution steps

\int \frac{cos ( ln ( x ) ) sin ( ln ( x ) )}{x} d x

Rewrite using trig identities

= \int \frac{sin ( 2 ln ( x ) )}{2 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{sin ( 2 ln ( x ) )}{x} d x

Apply u-substitution

: \int \frac{sin ( u )}{2} d u

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

Take the constant out:

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

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

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{2} \cdot \frac{1}{2} (- cos (u))

Substitute back

u = 2 ln (x)

= \frac{1}{2} \cdot \frac{1}{2} (- cos (2 ln (x)))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- cos (2 ln (x))) : - \frac{1}{4} cos (2 ln (x))

= - \frac{1}{4} cos (2 ln (x))

Add a constant to the solution

= - \frac{1}{4} cos (2 ln (x)) + C

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What is the integral of (cos(ln(x))sin(ln(x)))/x ?
The integral of (cos(ln(x))sin(ln(x)))/x is -1/4 cos(2ln(x))+C