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प्रश्न
` ∫ x tan ^2 x dx
बेरीज
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उत्तर
\[\int x \tan^2 x\ dx\]
= ∫ x (sec2 x – 1) dx
\[= \int x_I . \sec_{II} ^2 \text{ x dx }- \int \text{ x dx }\]
\[ = x\int \sec^2 x - \int\left\{ \frac{d}{dx}\left( x \right)\int \sec^2 \text{ x dx } \right\}dx - \frac{x^2}{2} + C_1 \]
\[ = x . \tan x - \int1 . \text{ tan x dx } - \frac{x^2}{2} + C_1 \]
\[ = x \tan x - \text{ log }\left| \sec x \right| - \frac{x^2}{2} + C_1 + C_2 \]
` = x tan - log | sec x | - x^2/2 + C_1 + C_2 ( where C = C_1 + C_2 )`
\[ = x\int \sec^2 x - \int\left\{ \frac{d}{dx}\left( x \right)\int \sec^2 \text{ x dx } \right\}dx - \frac{x^2}{2} + C_1 \]
\[ = x . \tan x - \int1 . \text{ tan x dx } - \frac{x^2}{2} + C_1 \]
\[ = x \tan x - \text{ log }\left| \sec x \right| - \frac{x^2}{2} + C_1 + C_2 \]
` = x tan - log | sec x | - x^2/2 + C_1 + C_2 ( where C = C_1 + C_2 )`
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