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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{cos ( ln ( 4 u ) )}{2 u} d u

= \int \frac{cos ( ln ( 4 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{cos ( ln ( 4 u ) )}{u} d u

Apply u-substitution

: \int \frac{cos ( ln ( v ) )}{v} d v

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

Apply u-substitution

: \int cos (w) d w

= \frac{1}{2} \cdot \int cos (w) d w

Use the common integral:

\int cos (w) d w = sin (w)

= \frac{1}{2} sin (w)

Substitute back

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

Simplify

ln (4 x^{2}) : 2 ln (2 x)

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

Add a constant to the solution

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

6 y^{^{'}} = y^{2}, y (0) = 5

(2 x - 1)^{^{'}}

\int \frac{x ^{2}}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

y = e^{x} sin (x)

f (x) = 6 x^{6} (x^{4} - 6 x)

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