What is the integral of 48x^{-1/5}tan(x^{4/5}) ?

The integral of 48x^{-1/5}tan(x^{4/5}) is -60ln|cos(x^{4/5})|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 48x^{-1/5}tan(x^{4/5})

\int 48 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

- 60 ln cos (x^{\frac{4}{5}}) + C

Solution steps

\int 48 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

Take the constant out:

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

= 48 \cdot \int x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

Apply u-substitution

: \int \frac{5 tan ( u )}{4} d u

= 48 \cdot \int \frac{5 tan ( u )}{4} d u

Take the constant out:

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

= 48 \cdot \frac{5}{4} \cdot \int tan (u) d u

Rewrite using trig identities

= 48 \cdot \frac{5}{4} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= 48 \cdot \frac{5}{4} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= 48 \cdot \frac{5}{4} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 48 \cdot \frac{5}{4} (- ln ∣ v ∣)

Substitute back

= 48 \cdot \frac{5}{4} (- ln cos (x^{\frac{4}{5}}))

Simplify

48 \cdot \frac{5}{4} (- ln cos (x^{\frac{4}{5}})) : - 60 ln cos (x^{\frac{4}{5}})

= - 60 ln cos (x^{\frac{4}{5}})

Add a constant to the solution

= - 60 ln cos (x^{\frac{4}{5}}) + C

\int \frac{1}{( 1 + x ^{2} ) \cdot ln ( x + 1 + x ^{2} )} d x

\int (x^{2} - 2 x - 15) d x

\int 5 x^{4} e^{8 x^{5}} d x

\int - 4 x e^{- 4 x} d x

\int lo g_{7} (x) d x

What is the integral of 48x^{-1/5}tan(x^{4/5}) ?
The integral of 48x^{-1/5}tan(x^{4/5}) is -60ln|cos(x^{4/5})|+C