What is the integral of (cos(3x))/(sqrt(5+6sin(3x))) ?

The integral of (cos(3x))/(sqrt(5+6sin(3x))) is 1/9 sqrt(5+6sin(3x))+C

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integral of (cos(3x))/(sqrt(5+6sin(3x)))

\int \frac{cos ( 3 x )}{5 + 6 sin ( 3 x )} d x

\frac{1}{9} 5 + 6 sin (3 x) + C

Solution steps

\int \frac{cos ( 3 x )}{5 + 6 sin ( 3 x )} d x

Apply u-substitution

: \int \frac{cos ( u )}{3 5 + 6 sin ( u )} d u

= \int \frac{cos ( u )}{3 5 + 6 sin ( 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{cos ( u )}{5 + 6 sin ( u )} d u

Apply u-substitution

: \int \frac{1}{6 v} d v

= \frac{1}{3} \cdot \int \frac{1}{6 v} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{6} \cdot \int \frac{1}{v} d v

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{1}{6} \cdot 2 v^{\frac{1}{2}}

Substitute back

= \frac{1}{3} \cdot \frac{1}{6} \cdot 2 (5 + 6 sin (3 x))^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{6} \cdot 2 (5 + 6 sin (3 x))^{\frac{1}{2}} : \frac{1}{9} 5 + 6 sin (3 x)

= \frac{1}{9} 5 + 6 sin (3 x)

Add a constant to the solution

= \frac{1}{9} 5 + 6 sin (3 x) + C

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What is the integral of (cos(3x))/(sqrt(5+6sin(3x))) ?
The integral of (cos(3x))/(sqrt(5+6sin(3x))) is 1/9 sqrt(5+6sin(3x))+C