: Let I = 2 ( 7 - 4x )dx Put 7 - 4x = t - 4dx = dt I = - 4 2 t dt = - 4 t + C = - 4 ( 7 - 4x ) + C

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.2,Q.22,Page 305

Question

${\sec ^2}\left( {7 - 4x} \right)$

Step-by-step Solution

: Let $I = \int {{{\sec }^2}\left( {7 - 4x} \right)dx}$

Put $7 - 4x = t$ $\Rightarrow$ $- 4dx = dt$

$\therefore$ $I = - \cfrac{1}{4}\int {{{\sec }^2}t} dt = - \cfrac{1}{4}\tan t + C = - \cfrac{1}{4}\tan \left( {7 - 4x} \right) + C$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.