What is the integral of x^2sec(4-3x^3) ?

The integral of x^2sec(4-3x^3) is -1/9 ln|tan(4-3x^3)+sec(4-3x^3)|+C

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integral of x^2sec(4-3x^3)

\int x^{2} sec (4 - 3 x^{3}) d x

- \frac{1}{9} ln tan (4 - 3 x^{3}) + sec (4 - 3 x^{3}) + C

Solution steps

\int x^{2} sec (4 - 3 x^{3}) d x

Apply u-substitution

= \int \frac{sec ( 4 - 3 u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (4 - 3 u) d u

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} (- \frac{1}{3} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= \frac{1}{3} (- \frac{1}{3} ln tan (4 - 3 x^{3}) + sec (4 - 3 x^{3}))

Simplify

\frac{1}{3} (- \frac{1}{3} ln tan (4 - 3 x^{3}) + sec (4 - 3 x^{3})) : - \frac{1}{9} ln tan (4 - 3 x^{3}) + sec (4 - 3 x^{3})

= - \frac{1}{9} ln tan (4 - 3 x^{3}) + sec (4 - 3 x^{3})

Add a constant to the solution

= - \frac{1}{9} ln tan (4 - 3 x^{3}) + sec (4 - 3 x^{3}) + C

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What is the integral of x^2sec(4-3x^3) ?
The integral of x^2sec(4-3x^3) is -1/9 ln|tan(4-3x^3)+sec(4-3x^3)|+C