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

The integral of (2tan(x))/(2+3cos(x)) is ln|3cos(x)+2|-ln(3|cos(x)|)+C

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

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

ln ∣ 3 cos (x) + 2 ∣ - ln (3 ∣ cos (x) ∣) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{tan ( x )}{2 + 3 cos ( x )} d x

Apply u-substitution

= 2 \cdot \int - \frac{1}{u ( 3 u + 2 )} d u

Take the constant out:

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

= 2 (- \int \frac{1}{u ( 3 u + 2 )} d u)

Apply u-substitution

= 2 (- \int \frac{1}{v ( v - 2 )} d v)

Complete the square

v (v - 2) : (v - 1)^{2} - 1

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

Apply u-substitution

= 2 (- \int \frac{1}{w ^{2} - 1} d w)

Factor

w^{2} - 1 : - (- w^{2} + 1)

= 2 (- \int \frac{1}{- ( - w ^{2} + 1 )} d w)

Take the constant out:

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

= 2 (- (- \int \frac{1}{- w ^{2} + 1} d w))

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 2 (- (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})))

Substitute back

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

Simplify

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

= ln ∣ 3 cos (x) + 2 ∣ - ln (3 ∣ cos (x) ∣)

Add a constant to the solution

= ln ∣ 3 cos (x) + 2 ∣ - ln (3 ∣ cos (x) ∣) + C

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What is the integral of (2tan(x))/(2+3cos(x)) ?
The integral of (2tan(x))/(2+3cos(x)) is ln|3cos(x)+2|-ln(3|cos(x)|)+C