What is the integration of $\int {{{\tan }^2}} x{\sec ^4}xdx$ ~~~~~[NCERT Exemp. Q. 7,Page 164 ]

Let $I = \int {{{\tan }^2}} x{\sec ^4}xdx$

Let’s put $\tan x = t \Rightarrow {\sec ^2}xdx = dt$

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{{{{\tan }^5}x}}{5} + \frac{{{{\tan }^3}x}}{3} + C$

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