: Let I = 2 x e c 2 x dx = 2 x 2 xdx = 2 xdx = 2 x - 1 )dx = x - x + C

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.1,Q.19,Page 299

Question

$\int {\cfrac{{{{\sec }^2}x}}{{\cos e{c^2}x}}dx}$

Step-by-step Solution

: Let$I = \int {\cfrac{{{{\sec }^2}x}}{{\cos e{c^2}x}}dx} = \int {\cfrac{1}{{{{\cos }^2}x}} \cdot {{\sin }^2}xdx}$

$= \int {{{\tan }^2}xdx} = \int {\left( {{{\sec }^2}x - 1} \right)dx = \tan x - x + C}$

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