What is the integral of (sec^2(6x))/(3tan^2(6x)+15) ?

The integral of (sec^2(6x))/(3tan^2(6x)+15) is 1/(18sqrt(5))arctan(1/(sqrt(5))tan(6x))+C

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integral of (sec^2(6x))/(3tan^2(6x)+15)

\int \frac{sec ^{2} ( 6 x )}{3 tan ^{2} ( 6 x ) + 15} d x

\frac{1}{18 5} arctan (\frac{1}{5} tan (6 x)) + C

Solution steps

\int \frac{sec ^{2} ( 6 x )}{3 tan ^{2} ( 6 x ) + 15} d x

Apply u-substitution

: \int \frac{sec ^{2} ( u )}{6 ( 3 tan ^{2} ( u ) + 15 )} d u

= \int \frac{sec ^{2} ( u )}{6 ( 3 tan ^{2} ( u ) + 15 )} 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 \frac{sec ^{2} ( u )}{3 tan ^{2} ( u ) + 15} d u

Apply u-substitution

: \int \frac{1}{3 v ^{2} + 15} d v

= \frac{1}{6} \cdot \int \frac{1}{3 v ^{2} + 15} d v

Apply Integral Substitution

= \frac{1}{6} \cdot \int \frac{1}{3 5 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{6} \cdot \frac{1}{3 5} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{6} \cdot \frac{1}{3 5} arctan (w)

Substitute back

= \frac{1}{6} \cdot \frac{1}{3 5} arctan (\frac{3}{15} tan (6 x))

Simplify

\frac{1}{6} \cdot \frac{1}{3 5} arctan (\frac{3}{15} tan (6 x)) : \frac{1}{18 5} arctan (\frac{1}{5} tan (6 x))

= \frac{1}{18 5} arctan (\frac{1}{5} tan (6 x))

Add a constant to the solution

= \frac{1}{18 5} arctan (\frac{1}{5} tan (6 x)) + C

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What is the integral of (sec^2(6x))/(3tan^2(6x)+15) ?
The integral of (sec^2(6x))/(3tan^2(6x)+15) is 1/(18sqrt(5))arctan(1/(sqrt(5))tan(6x))+C