What is the integral of (sin(2x))/(9+cos^2(2x)) ?

The integral of (sin(2x))/(9+cos^2(2x)) is -1/6 arctan(1/3 cos(2x))+C

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integral of (sin(2x))/(9+cos^2(2x))

\int \frac{sin ( 2 x )}{9 + cos ^{2} ( 2 x )} d x

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

Solution steps

\int \frac{sin ( 2 x )}{9 + cos ^{2} ( 2 x )} d x

Apply u-substitution

: \int \frac{sin ( u )}{2 ( 9 + cos ^{2} ( u ) )} d u

= \int \frac{sin ( u )}{2 ( 9 + 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}{2} \cdot \int \frac{sin ( u )}{9 + cos ^{2} ( u )} d u

Apply u-substitution

: \int - \frac{1}{9 + v ^{2}} d v

= \frac{1}{2} \cdot \int - \frac{1}{9 + v ^{2}} d v

Take the constant out:

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

= \frac{1}{2} (- \int \frac{1}{9 + v ^{2}} d v)

Apply Integral Substitution

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (sin(2x))/(9+cos^2(2x)) ?
The integral of (sin(2x))/(9+cos^2(2x)) is -1/6 arctan(1/3 cos(2x))+C