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प्रश्न
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
बेरीज
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उत्तर
\[\text{ Let I } = \int\left[ \tan\left( \log x \right) + \sec^2 \left( \log x \right) \right]dx\]
\[\text{ Put log x = t }\]
\[ \Rightarrow x = e^t \]
\[ \Rightarrow dx = e^t dt\]
\[ \text{ ∴ I }= \int\left( \tan t + \sec^2 t \right) e^t dt\]
\[\text{ Here,} f(t) = \tan t\]
\[ \Rightarrow f'(t) = \sec^2 t\]
` \text{ let e}^t \tan(t) = p `
\[\text{ Diff both sides w . r . t t }\]
\[ e^t \left[ \tan t + \sec^2 t \right] = \frac{dp}{dt}\]
\[ \Rightarrow e^t \left[ \tan t + \sec^2 t \right]dt = dp\]
\[ ∴ I = \int dp\]
\[ = p + C\]
\[ = e^t \tan t + C\]
\[ = x \text{ tan (log x) }+ C\]
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