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

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.2,Q.30,Page 305

Question

$\cfrac{{\sin x}}{{1 + \cos x}}$

Step-by-step Solution

: Let $I = \int {\cfrac{{\sin x}}{{1 + \cos x}}} dx$
Put $1 + \cos x = t$

$\Rightarrow$ $- \sin xdx = dt$

$\therefore$ $I = - \int {\cfrac{{dt}}{t} = - \log \left| t \right|} + C$

$= - \log \left| {1 + \cos x} \right| + C = \log \left( {\cfrac{1}{{\left| {1 + \cos x} \right|}}} \right) + C$

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