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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Substitute back

u = x^{3} - 3

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

Simplify

\frac{1}{3} \cdot \frac{1}{3} tan (x^{3} - 3) : \frac{1}{9} tan (x^{3} - 3)

= \frac{1}{9} tan (x^{3} - 3)

Add a constant to the solution

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

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