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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

= \int - \frac{1}{2 u ^{2} + 2} d u

Take the constant out:

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

= - \int \frac{1}{2 u ^{2} + 2} d u

\frac{1}{2 u ^{2} + 2} = \frac{1}{2} \frac{1}{u ^{2} + 1}

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

Use the common integral:

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

= - \frac{1}{2} arctan (u)

Substitute back

u = cos (x)

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

Add a constant to the solution

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

\int \frac{1}{( x + 1 ) ^{3} ( x - 2 )} d x

\int \frac{3}{3 x - 2} d x

\int cos (u) d u

\int \frac{x ^{2} - 1}{x - 2} d x

\int 1 - x - x^{2} d x

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