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

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

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integral of-9x^2sec(3x^3+1)tan(3x^3+1)

\int - 9 x^{2} sec (3 x^{3} + 1) tan (3 x^{3} + 1) d x

- sec (3 x^{3} + 1) + C

Solution steps

\int - 9 x^{2} sec (3 x^{3} + 1) tan (3 x^{3} + 1) 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 (3 x^{3} + 1) tan (3 x^{3} + 1) d x

Apply u-substitution

: \int \frac{u sec ( 3 u ^{\frac{3}{2}} + 1 ) tan ( 3 u ^{\frac{3}{2}} + 1 )}{2 x} d u

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int 1 d w

= - 9 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= - 9 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot 1 \cdot w

Substitute back

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

Simplify

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

= - sec (3 x^{3} + 1)

Add a constant to the solution

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

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