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

The integral of 1/(cos(θ)(1+sin(θ))) is 1/4 ln|1+sin(θ)|-1/(2(1+sin(θ)))-1/4 ln|sin(θ)-1|+C

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integral of 1/(cos(θ)(1+sin(θ)))

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

\frac{1}{4} ln ∣ 1 + sin (θ) ∣ - \frac{1}{2 ( 1 + sin ( θ ) )} - \frac{1}{4} ln ∣ sin (θ) - 1 ∣ + C

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply u-substitution

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

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

Factor

\frac{1}{v ( - v ^{2} + 2 v )}

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{v ^{2} ( v - 2 )} : - \frac{1}{4 v} - \frac{1}{2 v ^{2}} + \frac{1}{4 ( v - 2 )}

= - \int - \frac{1}{4 v} - \frac{1}{2 v ^{2}} + \frac{1}{4 ( v - 2 )} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int \frac{1}{4 v} d v - \int \frac{1}{2 v ^{2}} d v + \int \frac{1}{4 ( v - 2 )} d v)

\int \frac{1}{4 v} d v = \frac{1}{4} ln ∣ v ∣

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

\int \frac{1}{4 ( v - 2 )} d v = \frac{1}{4} ln ∣ v - 2 ∣

= - (- \frac{1}{4} ln ∣ v ∣ - (- \frac{1}{2 v}) + \frac{1}{4} ln ∣ v - 2 ∣)

Substitute back

= - (- \frac{1}{4} ln ∣ 1 + sin (θ) ∣ - (- \frac{1}{2 ( 1 + sin ( θ ) )}) + \frac{1}{4} ln ∣ 1 + sin (θ) - 2 ∣)

Simplify

- (- \frac{1}{4} ln ∣ 1 + sin (θ) ∣ - (- \frac{1}{2 ( 1 + sin ( θ ) )}) + \frac{1}{4} ln ∣ 1 + sin (θ) - 2 ∣) : \frac{1}{4} ln ∣ 1 + sin (θ) ∣ - \frac{1}{2 ( 1 + sin ( θ ) )} - \frac{1}{4} ln ∣ sin (θ) - 1 ∣

= \frac{1}{4} ln ∣ 1 + sin (θ) ∣ - \frac{1}{2 ( 1 + sin ( θ ) )} - \frac{1}{4} ln ∣ sin (θ) - 1 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 1 + sin (θ) ∣ - \frac{1}{2 ( 1 + sin ( θ ) )} - \frac{1}{4} ln ∣ sin (θ) - 1 ∣ + C

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What is the integral of 1/(cos(θ)(1+sin(θ))) ?
The integral of 1/(cos(θ)(1+sin(θ))) is 1/4 ln|1+sin(θ)|-1/(2(1+sin(θ)))-1/4 ln|sin(θ)-1|+C