What is the integral of (sin(3x))/(cos^2(3x)+6) ?

The integral of (sin(3x))/(cos^2(3x)+6) is -1/(3sqrt(6))arctan(1/(sqrt(6))cos(3x))+C

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integral of (sin(3x))/(cos^2(3x)+6)

\int \frac{sin ( 3 x )}{cos ^{2} ( 3 x ) + 6} d x

- \frac{1}{3 6} arctan (\frac{1}{6} cos (3 x)) + C

Solution steps

\int \frac{sin ( 3 x )}{cos ^{2} ( 3 x ) + 6} d x

Apply u-substitution

: \int \frac{sin ( u )}{3 ( cos ^{2} ( u ) + 6 )} d u

= \int \frac{sin ( u )}{3 ( cos ^{2} ( u ) + 6 )} 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 \frac{sin ( u )}{cos ^{2} ( u ) + 6} d u

Apply u-substitution

: \int - \frac{1}{v ^{2} + 6} d v

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

Take the constant out:

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

= \frac{1}{3} (- \int \frac{1}{v ^{2} + 6} d v)

Apply Integral Substitution

= \frac{1}{3} (- \int \frac{1}{6 ( 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}{3} (- \frac{1}{6} \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}{3} (- \frac{1}{6} arctan (w))

Substitute back

= \frac{1}{3} (- \frac{1}{6} arctan (\frac{cos ( 3 x )}{6}))

Simplify

\frac{1}{3} (- \frac{1}{6} arctan (\frac{cos ( 3 x )}{6})) : - \frac{1}{3 6} arctan (\frac{1}{6} cos (3 x))

= - \frac{1}{3 6} arctan (\frac{1}{6} cos (3 x))

Add a constant to the solution

= - \frac{1}{3 6} arctan (\frac{1}{6} cos (3 x)) + C

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What is the integral of (sin(3x))/(cos^2(3x)+6) ?
The integral of (sin(3x))/(cos^2(3x)+6) is -1/(3sqrt(6))arctan(1/(sqrt(6))cos(3x))+C