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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - \frac{1}{v} d v

= \frac{1}{3} \cdot \int - \frac{1}{v} 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} d v)

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= \frac{1}{3} (- 2 v^{\frac{1}{2}})

Substitute back

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

Simplify

\frac{1}{3} (- 2 (2 + cos (3 x))^{\frac{1}{2}}) : - \frac{2}{3} 2 + cos (3 x)

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

Add a constant to the solution

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

\int a r cos (x) d x

y = x \frac{a ^{2} - x ^{2}}{x}

x = y^{2} - 12, x = 4 y

\int cos (y) e^{5 s i n (y)} d y

n = 0 \sum \infty \frac{1 ^{n}}{n !}

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