Question

Q.Intigrate
ʃdx/a+bsinx

Answer 1

Answer on Question #53070 - Math - Integral Calculus

Integrate:

$$\int \frac {d x}{a + b \sin x}$$

Solution

Let

$$t = \tan \frac {x}{2}$$

By the double-angle formula for the sine function

$$\sin x = 2 \sin \frac {x}{2} \cos \frac {x}{2} = 2 t \cos^ {2} \frac {x}{2} = \frac {2 t}{\sec^ {2} \frac {x}{2}} = \frac {2 t}{1 + t ^ {2}}$$

The differential $dx$ can be calculated as follows:

$$\frac {d t}{d x} = \frac {1}{2} \sec^ {2} \frac {x}{2} = \frac {1 + t ^ {2}}{2}$$ $$d x = \frac {2}{1 + t ^ {2}} d t$$

Substitute

$$\begin{array}{l} \int \frac {d x}{a + b \sin x} = \int \frac {\frac {2}{1 + t ^ {2}} d t}{a + b \frac {2 t}{1 + t ^ {2}}} = \int \frac {\frac {2}{1 + t ^ {2}} d t}{\frac {a + a t ^ {2} + 2 b t}{1 + t ^ {2}}} = \int \frac {2 d t}{a + a t ^ {2} + 2 b t} \\ = \int \frac {2 d t}{a \left(t ^ {2} + \frac {2 b}{a} t\right) + a} = \int \frac {2 d t}{a \left(t ^ {2} + 2 \frac {b}{a} t + \left(\frac {b}{a}\right) ^ {2} - \left(\frac {b}{a}\right) ^ {2}\right) + a} \\ = \int \frac {2 d t}{a \left(\left(t + \frac {b}{a}\right) ^ {2} - \left(\frac {b}{a}\right) ^ {2}\right) + a} \\ = \int \frac {2 d t}{a \left(t + \frac {b}{a}\right) ^ {2} - \frac {b ^ {2}}{a} + a} = \frac {2}{a} \int \frac {d t}{\left(t + \frac {b}{a}\right) ^ {2} + 1 - \left(\frac {b}{a}\right) ^ {2}} \\ = \frac {2}{a} \int \frac {d (t + \frac {b}{a})}{(t + \frac {b}{a}) ^ {2} + 1 - (\frac {b}{a}) ^ {2}} = \frac {2}{a} \int \frac {d (t + \frac {b}{a})}{(t + \frac {b}{a}) ^ {2} + 1 - (\frac {b}{a}) ^ {2}} \\ \end{array}$$

Now we have 3 different cases:

1) If $1 - \left(\frac{b}{a}\right)^2 \geq 0$ , which is equivalent to $a^2 \geq b^2$

$$\begin{array}{l} I = \frac{2}{a} \int \frac{d \left(t + \frac{b}{a}\right)}{\left(t + \frac{b}{a}\right)^{2} + 1 - \left(\frac{b}{a}\right)^{2}} = \frac{2}{a} \frac{1}{\sqrt{1 - \left(\frac{b}{a}\right)^{2}}} \arctan \left(\frac{t + \frac{b}{a}}{\sqrt{1 - \left(\frac{b}{a}\right)^{2}}}\right) + C = \frac{2}{\sqrt{a^{2} - b^{2}}} \arctan \left(\frac{a \tan \frac{x}{2} + b}{\sqrt{a^{2} - b^{2}}}\right) + \\ + C, \end{array}$$

where $C$ is an arbitrary real constant.

2) If $1 - \left(\frac{b}{a}\right)^{2} = 0 \quad \Rightarrow \quad a^{2} = b^{2}$

$$\begin{array}{l} I = \frac{2}{a} \int \frac{d \left(t + \frac{b}{a}\right)}{\left(t + \frac{b}{a}\right)^{2} + 1 - \left(\frac{b}{a}\right)^{2}} = \frac{2}{a} \int \frac{d \left(t + \frac{b}{a}\right)}{\left(t + \frac{b}{a}\right)^{2}} = \frac{2}{a} \int \left(t + \frac{b}{a}\right)^{-2} d \left(t + \frac{b}{a}\right) = \frac{2}{a} \frac{\left(t + \frac{b}{a}\right)^{-2+1}}{(-2+1)} \\ + C = \frac{-2}{a \left(t + \frac{b}{a}\right)} + C = \frac{-2}{a \tan \frac{x}{2} + b} + C \end{array}$$

3) If $1 - \left(\frac{b}{a}\right)^{2} \leq 0 \quad \Rightarrow a^{2} \leq b^{2}$

$$\begin{array}{l} I = \frac{2}{a} \int \frac{d \left(t + \frac{b}{a}\right)}{\left(t + \frac{b}{a}\right)^{2} + 1 - \left(\frac{b}{a}\right)^{2}} = \frac{2}{a} \frac{1}{2} \ln \left| \frac{t + \frac{b}{a} - \left(\left(\frac{b}{a}\right)^{2} - 1\right)}{t + \frac{b}{a} + \left(\left(\frac{b}{a}\right)^{2} - 1\right)} \right| + C \\ = \frac{1}{a} \ln \left| \frac{\tan \frac{x}{2} + \frac{b}{a} - \left(\left(\frac{b}{a}\right)^{2} - 1\right)}{\tan \frac{x}{2} + \frac{b}{a} + \left(\left(\frac{b}{a}\right)^{2} - 1\right)} \right| + C \end{array}$$

Answer: $\frac{2}{\sqrt{a^{2} - b^{2}}} \arctan \left(\frac{a \tan \frac{x}{2} + b}{\sqrt{a^{2} - b^{2}}}\right) + C$, if $a^{2} > b^{2}$

$$\begin{array}{l} \frac{-2}{a \tan \frac{x}{2} + b} + C, \text{ if } a^{2} = b^{2} \\ \frac{1}{a} \ln \left| \frac{\tan \frac{x}{2} + \frac{b}{a} - \left(\left(\frac{b}{a}\right)^{2} - 1\right)}{\tan \frac{x}{2} + \frac{b}{a} + \left(\left(\frac{b}{a}\right)^{2} - 1\right)} \right| + C, \text{ if } a^{2} < b^{2} \end{array}$$

www.AssignmentExpert.com