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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u sec ( u ^{\frac{3}{2}} ) tan ( u ^{\frac{3}{2}} )}{2 x} 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})

Simplify

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

Add a constant to the solution

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

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