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