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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u sec ( u ^{\frac{3}{2}} - 1 )}{2 x} d u

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

Apply u-substitution

: \int sec (v) d v

= \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} ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Add a constant to the solution

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

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