Question

\[\int\frac{1}{a + b \tan x} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int\frac{1}{a + b \tan x}dx\]

\[ = \int\frac{1}{a + b \frac{\sin x}{\cos x}}dx\]

\[ = \int\frac{\cos x \cdot}{a \cos x + b \sin x}dx\]

\[\text{ Let } \cos x = \text{ A }\frac{d}{dx} \left( a \cos x + b \sin x \right) + \text{ B }\left( a \cos x + b \sin x \right)\]

\[ \Rightarrow \cos x = A \left( - a \sin x + b \cos x \right) + B \left( a \cos x + b \sin x \right)\]

\[1 \cdot \cos x = \left( Ab + B \cdot a \right) \cos x + \sin x\left( - A \cdot a + B \cdot b \right)\]

\[\text{Equating coefficients of like terms}\]

\[ A \cdot b + B \cdot a = 1 . . . \left( 1 \right)\]

\[ - A \cdot a + B \cdot b = 0 . . . \left( 2 \right)\]

\[\text{Multiplying equation} \left( 1 \right) \text{by a and eq} \left( 2 \right) \text{by b and then adding them} \]

\[ A \cdot ab + B \cdot a^2 = a\]

\[ - A \cdot a \cdot b + B b^2 = 0\]

\[ \Rightarrow B = \frac{a}{a^2 + b^2}\]

\[\text{Substituting the value of B in eq} \left( 1 \right)\]

\[ \Rightarrow A \cdot b + \frac{a^2}{a^2 + b^2} = 1\]

\[ \Rightarrow A \cdot b = 1 - \frac{a^2}{a^2 + b^2}\]

\[ \Rightarrow A = \frac{b}{a^2 + b^2}\]

\[ \therefore I = \frac{b}{a^2 + b^2}\int\left( \frac{- a \sin x + b \cos x}{a \cos x + b \sin x} \right)dx + \frac{a}{a^2 + b^2}\int\left( \frac{a \cos x + b \sin x}{a \cos x + b \sin x} \right)dx\]

\[ = \frac{b}{a^2 + b^2}\int\left( \frac{- a \sin x + b \cos x}{a \cos x + b \sin x} \right)dx + \frac{a}{a^2 + b^2}\int dx\]

\[\text{ Putting  a   cos x + b sin x = t in  the Ist  integral}\]

\[ \Rightarrow \left( - a \sin x + b \cos x \right)dx = dt\]

\[ \therefore I = \frac{b}{a^2 + b^2}\int\frac{dt}{t} + \frac{a}{a^2 + b^2}\int dx\]

\[ = \frac{b}{a^2 + b^2} \text{ ln }\left| t \right| + \frac{ax}{a^2 + b^2} + C\]

\[ = \frac{b}{a^2 + b^2} \text{ ln} \left| a \cos x + b \sin x \right| + \frac{ax}{a^2 + b^2} + C................ \left[ \because t = a \cos x + b \sin x \right]\]