What is the integral of 15(tan(x))(ln(cos(x))) ?

The integral of 15(tan(x))(ln(cos(x))) is -15/2 ln^2(cos(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 15(tan(x))(ln(cos(x)))

\int 15 (tan (x)) (ln (cos (x))) d x

- \frac{15}{2} ln^{2} (cos (x)) + C

Solution steps

\int 15 tan (x) ln (cos (x)) d x

Take the constant out:

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

= 15 \cdot \int tan (x) ln (cos (x)) d x

Apply u-substitution

: \int - \frac{ln ( u )}{u} d u

= 15 \cdot \int - \frac{ln ( u )}{u} d u

Take the constant out:

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

= 15 (- \int \frac{ln ( u )}{u} d u)

Apply u-substitution

: \int v d v

= 15 (- \int v d v)

Apply the Power Rule

: \frac{v ^{2}}{2}

= 15 (- \frac{v ^{2}}{2})

Substitute back

= 15 (- \frac{ln ^{2} ( cos ( x ) )}{2})

Simplify

15 (- \frac{ln ^{2} ( cos ( x ) )}{2}) : - \frac{15}{2} ln^{2} (cos (x))

= - \frac{15}{2} ln^{2} (cos (x))

Add a constant to the solution

= - \frac{15}{2} ln^{2} (cos (x)) + C

\int \frac{2 x}{x ^{2} - 4} d x

\int - cos (x) + 1 d x

x \to 0 lim (\frac{( x ^{2} + 2 x )}{x})

\frac{\partial}{\partial y} (- 2 x y (- 15 - x - y))

\int_{0}^{1} \frac{2}{x ( x + 1 )} d x

What is the integral of 15(tan(x))(ln(cos(x))) ?
The integral of 15(tan(x))(ln(cos(x))) is -15/2 ln^2(cos(x))+C