Let I = 2 x 4 xdx Let's put x = t xdx = dt therefore, I = ( 1 + t 2 )dt = + t 4 ) dt = 3 + 5 + C = 5 x 5 + 3 x 3 + C

Integrals — Class 12 Maths Solution

exemplar sa SA NCERT Exemp. Q. 7,Page 164

Question

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

Step-by-step Solution

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

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

therefore,$I = \int {{t^2}} \left( {1 + {t^2}} \right)dt = \int {\left( {{t^2} + {t^4}} \right)} dt$

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

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