What is the integral of (sin(ln(4x^2)))/x ?

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

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integral of (sin(ln(4x^2)))/x

\int \frac{sin ( ln ( 4 x ^{2} ) )}{x} d x

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

Solution steps

\int \frac{sin ( ln ( 4 x ^{2} ) )}{x} d x

Apply u-substitution

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

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

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 ( ln ( u ) )}{u} d u

Apply u-substitution

: \int sin (v) d v

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

Use the common integral:

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

= \frac{1}{2} (- cos (v))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x \to 1 lim (- 2 1 - x + 2)

\frac{d}{d x} (sin (x^{2} + 2 x))

x \to 0 lim (cos (x) ln (x))

f (x) = 4 sin (2 x)

x \to \infty lim (- 5)

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