What is the integral of 9x^2sec(x^3)tan(x^3) ?

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

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integral of 9x^2sec(x^3)tan(x^3)

\int 9 x^{2} sec (x^{3}) tan (x^{3}) d x

3 sec (x^{3}) + C

Solution steps

\int 9 x^{2} sec (x^{3}) tan (x^{3}) d x

Take the constant out:

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

= 9 \cdot \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

= 9 \cdot \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

= 9 \cdot \frac{1}{3} \cdot \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= 9 \cdot \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 9 \cdot \frac{1}{3} \cdot 1 \cdot v

Substitute back

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

Simplify

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

= 3 sec (x^{3})

Add a constant to the solution

= 3 sec (x^{3}) + C

sec (cos (x))

\int 2 x^{2} + x d x

\frac{d}{d x} (11 arctan (x))

x \to 5 + lim (\frac{( x + 1 )}{x - 5})

3 x^{2} - 2 x

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