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

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

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integral of sec(3x+2)tan(3x+2)

integral

sec (3 x + 2) tan (3 x + 2)

\frac{1}{3} sec (3 x + 2) + C

Solution steps

\int sec (3 x + 2) tan (3 x + 2) d x

Apply u-substitution

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

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

Apply u-substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

= \frac{1}{3} \cdot 1 \cdot sec (3 x + 2)

Simplify

= \frac{1}{3} sec (3 x + 2)

Add a constant to the solution

= \frac{1}{3} sec (3 x + 2) + C

x^{7} - 2 x^{5} + 2 x^{5} + x - 1

x^{\frac{1}{8}}

(2 x - 1) e^{2 x}

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

3 y^{2}

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