What is the integral of 6x^4tan(3x^5) ?

The integral of 6x^4tan(3x^5) is -2/5 ln|cos(3x^5)|+C

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integral of 6x^4tan(3x^5)

\int 6 x^{4} tan (3 x^{5}) d x

- \frac{2}{5} ln cos (3 x^{5}) + C

Solution steps

\int 6 x^{4} tan (3 x^{5}) d x

Take the constant out:

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

= 6 \cdot \int x^{4} tan (3 x^{5}) d x

Apply u-substitution

: \int \frac{tan ( u )}{15} d u

= 6 \cdot \int \frac{tan ( u )}{15} d u

Take the constant out:

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

= 6 \cdot \frac{1}{15} \cdot \int tan (u) d u

Rewrite using trig identities

= 6 \cdot \frac{1}{15} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= 6 \cdot \frac{1}{15} \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

= 6 \cdot \frac{1}{15} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 6 \cdot \frac{1}{15} (- ln ∣ v ∣)

Substitute back

= 6 \cdot \frac{1}{15} (- ln cos (3 x^{5}))

Simplify

6 \cdot \frac{1}{15} (- ln cos (3 x^{5})) : - \frac{2}{5} ln cos (3 x^{5})

= - \frac{2}{5} ln cos (3 x^{5})

Add a constant to the solution

= - \frac{2}{5} ln cos (3 x^{5}) + C

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What is the integral of 6x^4tan(3x^5) ?
The integral of 6x^4tan(3x^5) is -2/5 ln|cos(3x^5)|+C