What is the integral of 6x^2tan(x^3) ?

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

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integral of 6x^2tan(x^3)

\int 6 x^{2} tan (x^{3}) d x

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

Solution steps

\int 6 x^{2} tan (x^{3}) d x

Take the constant out:

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

= 6 \cdot \int x^{2} tan (x^{3}) d x

Apply u-substitution

: \int \frac{tan ( u )}{3} d u

= 6 \cdot \int \frac{tan ( u )}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int tan (u) d u

Rewrite using trig identities

= 6 \cdot \frac{1}{3} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= 6 \cdot \frac{1}{3} \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}{3} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 6 \cdot \frac{1}{3} (- ln ∣ v ∣)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x}, 16 x^{2} - 16 x y + y^{2} = 1

f (x) = (x^{3} - 9 x)^{8}, (- 3, 0)

\frac{d y}{d x} + 3 y = 2 x

x^{3}, x, 0, 1

x \to 4 lim (\frac{9}{9 - ( - 3 )})

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