What is the integral of 16x^{-1/5}tan(x^{4/5}) ?

The integral of 16x^{-1/5}tan(x^{4/5}) is -20ln|cos(x^{4/5})|+C

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integral of 16x^{-1/5}tan(x^{4/5})

\int 16 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

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

Solution steps

\int 16 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

= 16 \cdot \int x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

Apply u-substitution

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

= 16 \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

= 16 \cdot \frac{5}{4} \cdot \int tan (u) d u

Rewrite using trig identities

= 16 \cdot \frac{5}{4} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= 16 \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

= 16 \cdot \frac{5}{4} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 16 \cdot \frac{5}{4} (- ln ∣ v ∣)

Substitute back

= 16 \cdot \frac{5}{4} (- ln cos (x^{\frac{4}{5}}))

Simplify

16 \cdot \frac{5}{4} (- ln cos (x^{\frac{4}{5}})) : - 20 ln cos (x^{\frac{4}{5}})

= - 20 ln cos (x^{\frac{4}{5}})

Add a constant to the solution

= - 20 ln cos (x^{\frac{4}{5}}) + C

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What is the integral of 16x^{-1/5}tan(x^{4/5}) ?
The integral of 16x^{-1/5}tan(x^{4/5}) is -20ln|cos(x^{4/5})|+C