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प्रश्न
\[\int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right) dx\]
योग
उत्तर
\[\text{We have}, \]
\[I = \int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right)dx\]
\[ = \int e^{2x} \left( \frac{1}{1 + \cos 2x} + \frac{\sin 2x}{1 + \cos 2x} \right)dx\]
\[ = \int e^{2x} \left( \frac{1}{2 \cos^2 x} + \frac{2 \sin x \cos x}{2 \cos^2 x} \right)dx\]
\[ = \int e^{2x} \left( \frac{\sec^2 x}{2} + \tan x \right)dx\]
\[\text{ Let e}^{2x} \tan x = t\]
\[ \Rightarrow \left( e^{2x} \sec^2 x + 2 e^{2x} \tan x \right)dx = dt\]
\[ = \left[ \frac{e^{2x} \sec^2 x}{2} + e^{2x} \tan x \right]dx = \frac{dt}{2}\]
\[ \therefore I = \int \frac{dt}{2}\]
\[ = \frac{t}{2} + C\]
\[ = \frac{e^{2x} \tan x}{2} + C\]
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