What is the integral of ((cos(3x)))/(1+sin(3x)) ?

The integral of ((cos(3x)))/(1+sin(3x)) is 1/3 ln|1+sin(3x)|+C

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integral of ((cos(3x)))/(1+sin(3x))

\int \frac{( cos ( 3 x ) )}{1 + sin ( 3 x )} d x

\frac{1}{3} ln ∣ 1 + sin (3 x) ∣ + C

Solution steps

\int \frac{cos ( 3 x )}{1 + sin ( 3 x )} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 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{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + sin (3 x)

= \frac{1}{3} ln ∣ 1 + sin (3 x) ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 1 + sin (3 x) ∣ + C

\int_{0}^{1} 2 π x (12 - 12 x) d x

e^{4}

y^{^{''}} - \frac{9}{t} y^{^{'}} + \frac{34}{t ^{2}} y = 0

\frac{\partial}{\partial x} (\frac{1}{2 1 + x + y})

\frac{d y}{d x} = \frac{3 x + x y ^{2}}{y + x ^{2} y}

What is the integral of ((cos(3x)))/(1+sin(3x)) ?
The integral of ((cos(3x)))/(1+sin(3x)) is 1/3 ln|1+sin(3x)|+C