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प्रश्न
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
बेरीज
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उत्तर
\[\text{ Let I } = \int\frac{\sin^2 x}{\cos^6 x}dx\]
\[ = \int\frac{\sin^2 x}{\cos^2 x \cdot \cos^4 x}\text{ dx }\]
\[ = \int \tan^2 x \cdot \sec^4 \text{ x dx}\]
\[ = \int \tan^2 x \sec^2 x \cdot \sec^2 \text{ x dx}\]
\[ = \int \tan^2 x \left( 1 + \tan^2 x \right) \sec^2 \text{ x dx }\]
\[\text{ Putting tan x = t }\]
\[ \Rightarrow \sec^2 \text{ x dx = dt}\]
\[ \therefore I = \int t^2 \left( 1 + t^2 \right)dt\]
\[ = \int\left( t^2 + t^4 \right)dt\]
\[ = \frac{t^3}{3} + \frac{t^5}{5} + C\]
\[ = \frac{1}{3} \tan^3 x + \frac{1}{5} \tan^5 x + C............. \left[ \because t = \tan x \right]\]
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