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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{sec ( u ) tan ( 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 (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot sec (x^{3} - 1)

Simplify

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

Add a constant to the solution

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

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