Check the solution to "Help with ∫1+sin⁡x(1+2cos⁡x)sin⁡xdx"

Tony Mccarthy

Answered question

2022-03-24

Help with

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

Answer & Explanation

Cason Singleton

Beginner2022-03-25Added 13 answers

Trigonometric functions can be expressed using rational functions $t = \tan \frac{x}{2}$ .Here, specifically, we must
$\sin x = \frac{2 t}{1 + t^{2}}, \cos x = \frac{1 - t^{2}}{1 + t^{2}}$
So we proceed by substitution: if
$d t = \frac{1}{2} (1 + \tan^{2} \frac{x}{2}) d x \Leftrightarrow d x = \frac{2 d t}{1 + \tan^{2} x} and$
$\int \frac{1 + \sin x}{(1 + 2 \cos x) \sin x} d x = \int \frac{1 + \frac{2 t}{1 + t^{2}}}{(1 + \frac{2 (1 - t^{2})}{1 + t^{2}}) \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$
$= \int \frac{(1 + t^{2} + 2 t) 2 d t}{(1 + t^{2} + 2 (1 - t^{2})) t} = \int \frac{2 {(1 + t)}^{2}}{t (3 - t^{2})} d t$
The integral of a rational function is now in front of you, and it needs to be divided into partial fractions as follows:
$\frac{2 {(1 + t)}^{2}}{t (3 - t^{2})} = \frac{A}{t} + \frac{B t}{3 - t^{2}} (A, B \in R)$

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