What is the integral of sec^2(3x-4) ?

The integral of sec^2(3x-4) is 1/3 tan(3x-4)+C

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integral of sec^2(3x-4)

\int sec^{2} (3 x - 4) d x

\frac{1}{3} tan (3 x - 4) + C

Solution steps

\int sec^{2} (3 x - 4) d x

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = 3 x - 4

= \frac{1}{3} tan (3 x - 4)

Add a constant to the solution

= \frac{1}{3} tan (3 x - 4) + C

n = 0 \sum \infty n

\int 1 - cos^{2} (2 x) d x

R (x) = 71 x

\int \frac{3}{5 y y ^{2} - 1} d y

\frac{d}{d x} (- \frac{1 - x}{2})

What is the integral of sec^2(3x-4) ?
The integral of sec^2(3x-4) is 1/3 tan(3x-4)+C