What is the integral of (sin(3θ))/(8+cos^2(3θ)) ?

The integral of (sin(3θ))/(8+cos^2(3θ)) is -1/(6sqrt(2))arctan(1/(2sqrt(2))cos(3θ))+C

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integral of (sin(3θ))/(8+cos^2(3θ))

\int \frac{sin ( 3 θ )}{8 + cos ^{2} ( 3 θ )} d θ

- \frac{1}{6 2} arctan (\frac{1}{2 2} cos (3 θ)) + C

Solution steps

\int \frac{sin ( 3 θ )}{8 + cos ^{2} ( 3 θ )} d θ

Apply u-substitution

: \int \frac{sin ( u )}{3 ( 8 + cos ^{2} ( u ) )} d u

= \int \frac{sin ( u )}{3 ( 8 + cos ^{2} ( u ) )} 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 )}{8 + cos ^{2} ( u )} d u

Apply u-substitution

: \int - \frac{1}{8 + v ^{2}} d v

= \frac{1}{3} \cdot \int - \frac{1}{8 + v ^{2}} 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}{8 + v ^{2}} d v)

Apply Integral Substitution

= \frac{1}{3} (- \int \frac{1}{2 2 ( 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}{2 2} \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}{2 2} arctan (w))

Substitute back

= \frac{1}{3} (- \frac{1}{2 2} arctan (\frac{cos ( 3 θ )}{2 2}))

Simplify

\frac{1}{3} (- \frac{1}{2 2} arctan (\frac{cos ( 3 θ )}{2 2})) : - \frac{1}{6 2} arctan (\frac{1}{2 2} cos (3 θ))

= - \frac{1}{6 2} arctan (\frac{1}{2 2} cos (3 θ))

Add a constant to the solution

= - \frac{1}{6 2} arctan (\frac{1}{2 2} cos (3 θ)) + C

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What is the integral of (sin(3θ))/(8+cos^2(3θ)) ?
The integral of (sin(3θ))/(8+cos^2(3θ)) is -1/(6sqrt(2))arctan(1/(2sqrt(2))cos(3θ))+C