What is the integral of (-9x^2sec(3x^3+1)) ?

The integral of (-9x^2sec(3x^3+1)) is -ln|tan(1+3x^3)+sec(1+3x^3)|+C

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integral of (-9x^2sec(3x^3+1))

\int (- 9 x^{2} sec (3 x^{3} + 1)) d x

- ln tan (1 + 3 x^{3}) + sec (1 + 3 x^{3}) + C

Solution steps

\int - 9 x^{2} sec (3 x^{3} + 1) d x

Take the constant out:

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

= - 9 \cdot \int x^{2} sec (3 x^{3} + 1) d x

Apply u-substitution

: \int \frac{sec ( u )}{9} d u

= - 9 \cdot \int \frac{sec ( u )}{9} d u

Take the constant out:

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

= - 9 \cdot \frac{1}{9} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= - 9 \cdot \frac{1}{9} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = 1 + 3 x^{3}

= - 9 \cdot \frac{1}{9} ln tan (1 + 3 x^{3}) + sec (1 + 3 x^{3})

Simplify

- 9 \cdot \frac{1}{9} ln tan (1 + 3 x^{3}) + sec (1 + 3 x^{3}) : - ln tan (1 + 3 x^{3}) + sec (1 + 3 x^{3})

= - ln tan (1 + 3 x^{3}) + sec (1 + 3 x^{3})

Add a constant to the solution

= - ln tan (1 + 3 x^{3}) + sec (1 + 3 x^{3}) + C

\frac{d}{d x} ((4 x)^{5 x})

\int (3 x)^{2} + 2 d x

x \to 1 lim (1 - 2 - x)

\frac{d}{d x} (arccos (x^{3}))

\frac{d}{d x} (2 \cdot x^{2})

What is the integral of (-9x^2sec(3x^3+1)) ?
The integral of (-9x^2sec(3x^3+1)) is -ln|tan(1+3x^3)+sec(1+3x^3)|+C