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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int sec^{2} (v) \frac{1}{3} d v

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

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} (v) d v

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y = x^{5} ln (x)

x \to 0 lim (\frac{2 8 ^{x} - 1}{x})

(arcsin (5 x + 5))^{8}

x \to - 2 lim ((x + 4)^{3})

\frac{d}{d x} (30 x^{4} - 80 x^{3} + 60 x^{2} - 10)

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