What is the integral of (sec(θ)tan(θ))/(2sec(θ)+3) ?

The integral of (sec(θ)tan(θ))/(2sec(θ)+3) is 1/2 ln|3cos(θ)+2|-1/2 ln(3|cos(θ)|)+C

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integral of (sec(θ)tan(θ))/(2sec(θ)+3)

\int \frac{sec ( θ ) tan ( θ )}{2 sec ( θ ) + 3} d θ

\frac{1}{2} ln ∣ 3 cos (θ) + 2 ∣ - \frac{1}{2} ln (3 ∣ cos (θ) ∣) + C

Solution steps

\int \frac{sec ( θ ) tan ( θ )}{2 sec ( θ ) + 3} d θ

Rewrite using trig identities

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

Simplify

\frac{sin ( θ )}{cos ( θ ) ( 3 cos ( θ ) + 2 )} : \frac{tan ( θ )}{3 cos ( θ ) + 2}

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

Apply u-substitution

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

= \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

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

Apply u-substitution

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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{w ^{2} - 1} d w

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

Factor

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

= - \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

= - (- \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}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (sec(θ)tan(θ))/(2sec(θ)+3) ?
The integral of (sec(θ)tan(θ))/(2sec(θ)+3) is 1/2 ln|3cos(θ)+2|-1/2 ln(3|cos(θ)|)+C