Question

Q. Intigrate
ʃdx/a+bcosx

Answer 1

Answer on Question #53068 – Math – Integral Calculus

Integrate $\int \frac{dx}{a + b\cos x}$

Solution

$$I = \int \frac{dx}{a + b\cos x} = \left| \begin{array}{l} t = \tan \frac{x}{2} \\ dx = \frac{2dt}{1 + t^2} \\ \cos x = \frac{1 - t^2}{1 + t^2} \end{array} \right| = \int \frac{\frac{2dt}{1 + t^2}}{a + b \cdot \frac{1 - t^2}{1 + t^2}} = 2\int \frac{dt}{a(1 + t^2) + b(1 - t^2)} = 2\int \frac{dt}{(a + b) + (a - b)t^2}$$

This integral depends on the parameters $a$ and $b$. It's obvious that

$$I = I(a, b) = \left\{ \begin{array}{c} I_1 = 2\int \frac{dt}{2a}, \text{ if } b = a \\ I_2 = 2\int \frac{dt}{2a t^2}, \text{ if } b = -a \\ I_3 = 2\int \frac{dt}{(a + b) + (a - b)t^2}, \text{ if } b \neq a \text{ and } b \neq -a \end{array} \right.$$ $$I_1 = \int \frac{dx}{a + a\cos x} = \left| \begin{array}{l} t = \tan \frac{x}{2} \\ dx = \frac{2dt}{1 + t^2} \\ \cos x = \frac{1 - t^2}{1 + t^2} \end{array} \right| = 2\int \frac{dt}{a(1 + t^2) + a(1 - t^2)} = 2\int \frac{dt}{2a} = \frac{2}{2a}\int dt = \frac{1}{a}t + c = \frac{1}{a}\tan \frac{x}{2} + c$$ $$I_2 = \int \frac{dx}{a - a\cos x} = \left| \begin{array}{l} t = \tan \frac{x}{2} \\ dx = \frac{2dt}{1 + t^2} \\ \cos x = \frac{1 - t^2}{1 + t^2} \end{array} \right| = 2\int \frac{dt}{a(1 + t^2) - a(1 - t^2)} = 2\int \frac{dt}{2a t^2} = \frac{2}{2a}\int \frac{1}{t^2} dt = -\frac{1}{a} \cdot \frac{1}{t} + c = -\frac{1}{a} \cdot \frac{1}{\tan \frac{x}{2}} + c$$ $$\begin{array}{l} I_3 = \int \frac{dx}{a + b\cos x} = \left| \begin{array}{l} t = \tan \frac{x}{2} \\ dx = \frac{2dt}{1 + t^2} \\ \cos x = \frac{1 - t^2}{1 + t^2} \end{array} \right| = 2\int \frac{dt}{a(1 + t^2) + b(1 - t^2)} = 2\int \frac{dt}{(a + b) + (a - b)t^2} = \frac{2}{a - b}\int \frac{dt}{\frac{a + b}{a - b} + t^2} = \\ = \left\{ \begin{array}{l} I_{31}, \text{ if } a > b \\ I_{32}, \text{ if } a < b \end{array} \right. \\ \end{array}$$ $$I_{32} = \frac{2}{a - b} \int \frac{dt}{t^2 - \gamma^2} = \left| \gamma^2 = -\frac{(a + b)}{(a - b)} = \frac{(a + b)}{(b - a)} > 0 \right| \text{ To evaluate } I_{32} \text{ apply the Partial Fraction}$$

Decomposition Method.

$$\begin{array}{l} \frac{1}{t^{2} - \gamma^{2}} = \frac{A}{t - \gamma} + \frac{B}{t + \gamma} = \frac{A(t + \gamma) + B(t - \gamma)}{t^{2} - \gamma^{2}} = \frac{(A + B)t + (A - B)\gamma}{t^{2} - \gamma^{2}} \Rightarrow \\ \left\{ \begin{array}{l} A + B = 0 \\ A - B = \frac{1}{\gamma} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} A = \frac{1}{2\gamma} \\ B = -\frac{1}{2\gamma} \end{array} \right. \\ \frac{1}{t^{2} - \gamma^{2}} = \frac{1}{2\gamma} \frac{1}{t - \gamma} - \frac{1}{2\gamma} \frac{1}{t + \gamma} \\ I_{32} = \frac{2}{a - b} \int \frac{dt}{t^{2} - \gamma^{2}} = \frac{2}{a - b} \frac{1}{2\gamma} \int \frac{dt}{t - \gamma} - \frac{2}{a - b} \frac{1}{2\gamma} \int \frac{dt}{t + \gamma} = \frac{1}{a - b} \frac{1}{\gamma} \ln |t - \gamma| - \frac{1}{a - b} \frac{1}{\gamma} \ln |t + \gamma| + c = \\ = \frac{1}{a - b} \frac{1}{\gamma} \ln \left| \frac{t - \gamma}{|t + \gamma|} \right| + c = \frac{1}{a - b} \frac{1}{\sqrt{\frac{(a + b)}{(b - a)}}} \ln \left| \frac{\tan \frac{x}{2} - \sqrt{\frac{(a + b)}{(b - a)}}}{\tan \frac{x}{2} + \sqrt{\frac{(a + b)}{(b - a)}}} \right| + c = \\ = -\frac{1}{b - a} \sqrt{\frac{b - a}{a + b}} \ln \left| \frac{\tan \frac{x}{2} - \sqrt{\frac{(a + b)}{(b - a)}}}{\tan \frac{x}{2} + \sqrt{\frac{(a + b)}{(b - a)}}} \right| + c = -\frac{1}{\sqrt{b^{2} - a^{2}}} \ln \left| \frac{\tan \frac{x}{2} - \sqrt{\frac{(a + b)}{(b - a)}}}{\tan \frac{x}{2} + \sqrt{\frac{(a + b)}{(b - a)}}} \right| + c \\ \end{array}$$

$c$ is the integration constant.

$$\begin{array}{l} I_{31} = \frac{2}{a - b} \int \frac{dt}{\left( \sqrt{\frac{(a + b)}{(a - b)}} \right)^{2} + t^{2}} = \frac{2}{a - b} \sqrt{\frac{(a - b)}{(a + b)}} \arctan \left( t \sqrt{\frac{(a - b)}{(a + b)}} \right) + c = \\ = \frac{2}{a - b} \sqrt{\frac{(a - b)}{(a + b)}} \arctan \left( \sqrt{\frac{(a - b)}{(a + b)}} \tan \frac{x}{2} \right) + c = \\ = \frac{2}{\sqrt{(a - b)(a + b)}} \arctan \left( \sqrt{\frac{(a - b)}{(a + b)}} \tan \frac{x}{2} \right) + c = \frac{2}{\sqrt{(a^{2} - b^{2})}} \arctan \left( \sqrt{\frac{(a - b)}{(a + b)}} \tan \frac{x}{2} \right) + c \\ \end{array}$$

Answer: $I = I(a,b) = \left\{

\begin{array}{l}

\frac{1}{a} \tan \frac{x}{2} + c \text{ if } b = a \\

-\frac{1}{a} \frac{1}{\tan \frac{x}{2}} + c \text{ if } b = -a \\

\frac{2}{\sqrt{(a^{2} - b^{2})}} \arctan \left( \sqrt{\frac{(a - b)}{(a + b)}} \cdot \tan \frac{x}{2} \right) + c, \text{ if } a > b \\

-\frac{1}{\sqrt{b^{2} - a^{2}}} \ln \left| \frac{\tan \frac{x}{2} - \sqrt{\frac{(a + b)}{(b - a)}}}{\tan \frac{x}{2} + \sqrt{\frac{(a + b)}{(b - a)}}} \right| + c, \text{ if } a < b \\

\end{array}

\right.$

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