${\sec ^2}\left( {7 - 4x} \right)$
${\sec ^2}\left( {7 - 4x} \right)$
Official Solution
VVidaara Team
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NCERT & Exemplar
: Let $I = \int {{{\sec }^2}\left( {7 - 4x} \right)dx}$
Put $7 - 4x = t$ $\Rightarrow$ $- 4dx = dt$
$\therefore$ $I = - \cfrac{1}{4}\int {{{\sec }^2}t} dt = - \cfrac{1}{4}\tan t + C = - \cfrac{1}{4}\tan \left( {7 - 4x} \right) + C$
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