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प्रश्न
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उत्तर
\[\text{ Let I } = \int e^x \sec x\left( 1 + \tan x \right)dx\]
\[ = \int e^x \left( \sec x + \sec x \tan x \right)dx\]
\[\text{ Here, }f(x) = \text{ sec x Put e}^x f(x) = t\]
\[ \Rightarrow f'(x) = \sec x \tan x\]
\[\text{ let e}^x \sec x = t\]
\[\text{ Diff both sides w . r . t x }\]
\[ e^x \sec x + e^x \sec x \tan x = \frac{dt}{dx}\]
\[ \Rightarrow e^x \left( \sec x + \tan x \right)dx = dt\]
\[ \therefore \int e^x \left( \sec x + \sec x \tan x \right)dx = \int dt\]
\[ = t + C\]
\[ = e^x \sec x + C\]
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संबंधित प्रश्न
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int {cosec}^4 2x\ dx\]
