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

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

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

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

- \frac{1}{3} ln ∣ cos (3 x) + 1 ∣ + C

Solution steps

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

= - \frac{1}{3} ln ∣ cos (3 x) + 1 ∣

Add a constant to the solution

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

\int_{0}^{1} 2 x e^{x} d x

\int \frac{cos ^{2} ( x )}{1 + sin ( x )} d x

\int \frac{x}{( 3 x ^{2} + 8 ) ^{2}} d x

f (x) = ln (x + 3)

\int (3 x + 1) d x

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