What is the integral of-2x^4tan(x^5) ?

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

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integral of-2x^4tan(x^5)

\int - 2 x^{4} tan (x^{5}) d x

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

Solution steps

\int - 2 x^{4} tan (x^{5}) d x

Take the constant out:

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

= - 2 \cdot \int x^{4} tan (x^{5}) d x

Apply u-substitution

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

= - 2 \cdot \int \frac{tan ( u )}{5} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{5} \cdot \int tan (u) d u

Rewrite using trig identities

= - 2 \cdot \frac{1}{5} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= - 2 \cdot \frac{1}{5} \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

= - 2 \cdot \frac{1}{5} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - 2 \cdot \frac{1}{5} (- ln ∣ v ∣)

Substitute back

= - 2 \cdot \frac{1}{5} (- ln cos (x^{5}))

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (4 x^{3} y^{2} + 2)

x^{2} y^{^{''}} + x y^{^{'}} - 9 y = x^{3} + 1

n = 0 \sum \infty (1 + \frac{2}{n})^{3 n}

\frac{d}{d x} (x^{a} (1 - x)^{b})

x \to - 8 lim (\frac{- 5 x}{x + 8})

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