What is the integral of 1/(cos^2(3x)) ?

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

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integral of 1/(cos^2(3x))

\int \frac{1}{cos ^{2} ( 3 x )} d x

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

Solution steps

\int \frac{1}{cos ^{2} ( 3 x )} d x

Rewrite using trig identities

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

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

Add a constant to the solution

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

\int_{- 2}^{2} (x + 3) 4 - x^{2} d x

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

f (x) = 2 e^{x^{3}} + 1

f (x) = x + 3

\frac{d}{d x} (x + 5 e^{x})

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