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प्रश्न
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
बेरीज
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उत्तर
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} dx\]
Dividing numerator and denominator by cos2x we get ,
\[I = \int\frac{\frac{1}{\cos^2 x}}{\left( \tan x - 2 \right) \left( 2 \tan x + 1 \right)}dx\]
\[ = \int\frac{\sec^2 x}{\left( \tan x - 2 \right) \left( 2 \tan x + 1 \right)} dx\]
\[\text{ Putting tan x = t }\]
\[ \Rightarrow \text{ sec}^2 \text{ x dx} = dt\]
\[ \therefore I = \int\frac{1}{\left( t - 2 \right) \left( 2t + 1 \right)}dt\]
\[ = \int\frac{1}{2 t^2 + t - 4t - 2}dt\]
\[ = \int\frac{1}{2 t^2 - 3t - 2}dt\]
\[ = \frac{1}{2}\int\frac{1}{t^2 - \frac{3t}{2} - 1}dt\]
\[ = \frac{1}{2}\int\frac{1}{t^2 - \frac{3}{2}t + \left( \frac{3}{4} \right)^2 - \left( \frac{3}{4} \right)^2 - 1}dt\]
\[ = \frac{1}{2}\int\frac{1}{\left( t - \frac{3}{4} \right)^2 - \frac{9}{16} - 1}dt\]
\[ = \frac{1}{2}\int\frac{1}{\left( t - \frac{3}{4} \right)^2 - \left( \frac{5}{4} \right)^2}dt\]
\[ = \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \text{ ln }\left| \frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}} \right| + C ....................\left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]
\[ = \frac{1}{5} \text{ ln }\left| \frac{t - 2}{t + \frac{1}{2}} \right| + C\]
\[ = \frac{1}{5} \text{ ln } \left| \frac{2 \left( t - 2 \right)}{2t + 1} \right| + C\]
\[ = \frac{1}{5} \text{ ln }\left| \frac{2 \left( \tan x - 2 \right)}{2 \tan x + 1} \right| + C................ \left[ \because t = \tan x \right]\]
\[ = \frac{1}{5} \text{ ln} \left| \frac{\tan x - 2}{2 \tan x + 1} \right| + \frac{1}{5} \text{ ln 2 + C }\]
\[ = \frac{1}{5} \text{ ln }\left| \frac{\tan x - 2}{2 \tan x + 1} \right| + C'\]
\[\text{ where } \]
\[C' = C + \frac{1}{5} \text{ ln 2 }\]
\[ = \int\frac{\sec^2 x}{\left( \tan x - 2 \right) \left( 2 \tan x + 1 \right)} dx\]
\[\text{ Putting tan x = t }\]
\[ \Rightarrow \text{ sec}^2 \text{ x dx} = dt\]
\[ \therefore I = \int\frac{1}{\left( t - 2 \right) \left( 2t + 1 \right)}dt\]
\[ = \int\frac{1}{2 t^2 + t - 4t - 2}dt\]
\[ = \int\frac{1}{2 t^2 - 3t - 2}dt\]
\[ = \frac{1}{2}\int\frac{1}{t^2 - \frac{3t}{2} - 1}dt\]
\[ = \frac{1}{2}\int\frac{1}{t^2 - \frac{3}{2}t + \left( \frac{3}{4} \right)^2 - \left( \frac{3}{4} \right)^2 - 1}dt\]
\[ = \frac{1}{2}\int\frac{1}{\left( t - \frac{3}{4} \right)^2 - \frac{9}{16} - 1}dt\]
\[ = \frac{1}{2}\int\frac{1}{\left( t - \frac{3}{4} \right)^2 - \left( \frac{5}{4} \right)^2}dt\]
\[ = \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \text{ ln }\left| \frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}} \right| + C ....................\left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]
\[ = \frac{1}{5} \text{ ln }\left| \frac{t - 2}{t + \frac{1}{2}} \right| + C\]
\[ = \frac{1}{5} \text{ ln } \left| \frac{2 \left( t - 2 \right)}{2t + 1} \right| + C\]
\[ = \frac{1}{5} \text{ ln }\left| \frac{2 \left( \tan x - 2 \right)}{2 \tan x + 1} \right| + C................ \left[ \because t = \tan x \right]\]
\[ = \frac{1}{5} \text{ ln} \left| \frac{\tan x - 2}{2 \tan x + 1} \right| + \frac{1}{5} \text{ ln 2 + C }\]
\[ = \frac{1}{5} \text{ ln }\left| \frac{\tan x - 2}{2 \tan x + 1} \right| + C'\]
\[\text{ where } \]
\[C' = C + \frac{1}{5} \text{ ln 2 }\]
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