: Let I = 2 x 2 x + 4 dx Put x = t xdx = dt = + ( 2 ) 2 = | t + + 4 | + C = | x + 2 x + 4 | + C

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.4,Q.9,Page 315

Question

$\cfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}$

Step-by-step Solution

: Let $I = \cfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}dx$

Put $x = t$ $\Rightarrow$ ${\sec ^2}xdx = dt$

$= \int {\cfrac{{dt}}{{\sqrt {{t^2} + {{\left( 2 \right)}^2}} }} = \log \left| {t + \sqrt {{t^2} + 4} } \right| + C}$

$= \log \left| {\tan x + \sqrt {{{\tan }^2}x + 4} } \right| + C$

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