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प्रश्न
\[\int \sec^2 x \cos^2 2x \text{ dx }\]
बेरीज
उत्तर
\[\int\left( \sec^2 x \cdot \cos^2 2x \right)dx\]
\[ = \int \sec^2 x \times \left( 2 \cos^2 x - 1 \right)^2 dx\]
\[ = \int \sec^2 x \left[ 4 \cos^4 x - 4 \cos^2 x + 1 \right]dx\]
\[ \Rightarrow \int\left( 4 \cos^2 x - 4 + \sec^2 x \right)dx\]
\[ = 4\int \cos^2 x \text{ dx } + \int \sec^2 x \text{ dx }- 4\int dx\]
\[ \Rightarrow 4\int\left( \frac{1 + \cos 2x}{2} \right)dx + \int \sec^2 x - 4\int dx\]
\[ \Rightarrow 2 \left[ x + \frac{\sin 2x}{2} \right] + \tan x - 4x + C\]
\[ \Rightarrow \sin 2x + \tan x - 2x + C\]
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