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