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