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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{sec ^{2} ( 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^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= \frac{1}{3} tan (u)

Substitute back

u = x^{3}

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

Add a constant to the solution

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

x \to 0 lim (\frac{x ^{3} - 0}{x})

n = 1 \sum \infty n e^{- 3 n}

x \to \frac{π}{2} lim (\frac{1 - tan ( 2 x )}{2 tan ( x )})

x^{- 8 c o s (x)}

\frac{d}{d x} (4 x^{2} cos (2 x))

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