What is the integral of 1/(cos^2(6x)) ?

The integral of 1/(cos^2(6x)) is 1/6 tan(6x)+C

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integral of 1/(cos^2(6x))

\int \frac{1}{cos ^{2} ( 6 x )} d x

\frac{1}{6} tan (6 x) + C

Solution steps

\int \frac{1}{cos ^{2} ( 6 x )} d x

Rewrite using trig identities

= \int sec^{2} (6 x) d x

Apply u-substitution

: \int sec^{2} (u) \frac{1}{6} d u

= \int sec^{2} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{6} tan (u)

Substitute back

u = 6 x

= \frac{1}{6} tan (6 x)

Add a constant to the solution

= \frac{1}{6} tan (6 x) + C

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What is the integral of 1/(cos^2(6x)) ?
The integral of 1/(cos^2(6x)) is 1/6 tan(6x)+C