"Evaluating: ∫1+sin⁡(x)1+cos⁡(x)dx" was answered by students like you

Rolando Wade

Answered question

2022-04-02

Evaluating:

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

Ireland Vaughan

Beginner2022-04-03Added 14 answers

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

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

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

= \frac{1}{2} \int {(\tan (\frac{x}{2}) + 1)}^{2} d x

substitute

t = \tan \frac{x}{4}

now

\frac{1}{2} \int {(1 + \tan (\frac{x}{2}))}^{2} d x =

\int \frac{{(1 + t)}^{2}}{2} \times \frac{2 d t}{1 + t^{2}} =

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

t + \ln (1 + t^{2}) + C

Remark that you can substitute directly without intermediate trigonometric changes :

\frac{1 + \sin (x)}{1 + \cos (x)} = \frac{1 + \frac{2 t}{1 + t^{2}}}{1 + \frac{1 - t^{2}}{1 + t^{2}}} = \frac{1 + 2 t + t^{2}}{2}

Cassius Villarreal

Beginner2022-04-04Added 11 answers

An alternate way is this

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

= \int \csc^{2} (x) + \csc (x) - \csc (x) \cot (x) - \cot (x) d x

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