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प्रश्न
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
बेरीज
उत्तर
\[\text{ Let I } = \int\frac{1}{\sin^2 x + \sin 2x}dx\]
\[ = \int\frac{1}{\sin^2 x + 2 \sin x \cdot \cos x}dx\]
Dividing numerator and denominator by cos2x, we get
\[I = \int\frac{\frac{1}{\cos^2 x}}{\tan^2 x + 2 \tan x}dx\]
\[ = \int\frac{\sec^2 x}{\tan^2 x + 2 \tan x} dx\]
\[\text{ Putting tan x = t}\]
\[ \Rightarrow \text{ sec}^2 \text{ x dx = dt }\]
\[ \therefore I = \int\frac{1}{t^2 + 2t}dt\]
\[ = \int\frac{1}{t^2 + 2t + 1 - 1}dt\]
\[ = \int\frac{1}{\left( t + 1 \right)^2 - 1^2}dt\]
\[ = \frac{1}{2} \text{ ln} \left| \frac{t + 1 - 1}{t + 1 + 1} \right| + C\]
\[ = \frac{1}{2} \text{ ln } \left| \frac{t}{t + 2} \right| + C \]
\[ = \frac{1}{2} \text{ ln} \left| \frac{\tan x}{\tan x + 2} \right| + C ............\left[ \because t = \tan x \right]\]
\[ = \int\frac{\sec^2 x}{\tan^2 x + 2 \tan x} dx\]
\[\text{ Putting tan x = t}\]
\[ \Rightarrow \text{ sec}^2 \text{ x dx = dt }\]
\[ \therefore I = \int\frac{1}{t^2 + 2t}dt\]
\[ = \int\frac{1}{t^2 + 2t + 1 - 1}dt\]
\[ = \int\frac{1}{\left( t + 1 \right)^2 - 1^2}dt\]
\[ = \frac{1}{2} \text{ ln} \left| \frac{t + 1 - 1}{t + 1 + 1} \right| + C\]
\[ = \frac{1}{2} \text{ ln } \left| \frac{t}{t + 2} \right| + C \]
\[ = \frac{1}{2} \text{ ln} \left| \frac{\tan x}{\tan x + 2} \right| + C ............\left[ \because t = \tan x \right]\]
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