: Let I = ( 1 + x ) 2 dx Put 1 + x = t - xdx = dt I = - t 2 = - - 2 + 1 + C = t + C = 1 + x + C

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.2,Q.31,Page 305

Question

$\cfrac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}$

Step-by-step Solution

: Let$I = \int {\cfrac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}dx}$

Put $1 + \cos x = t$ $\Rightarrow$ $- \sin xdx = dt$

$\therefore$ $I = - \int {\cfrac{{dt}}{{{t^2}}} = - \cfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + C = \cfrac{1}{t} + C = \cfrac{1}{{1 + \cos x}} + C}$

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