Question

$\int \frac{dx}{(sinx+sin2x)}=$

• A. $\frac{1}{6}log(1-cosx)+\frac{1}{2}log(1+cosx)-\frac{2}{3}log(1+2cosx)$
  
• B. $6log(1-cosx)+2log(1+cosx)-\frac{2}{3}log(1+2cosx)$
  
• C. $6log(1-cosx)+\frac{1}{2}log(1+cosx)+\frac{2}{3}log(1+2cosx)$
  
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{1}{6}log(1-cosx)+\frac{1}{2}log(1+cosx)-\frac{2}{3}log(1+2cosx)$


$I=\int \frac{dx}{sinx(1+2cosx)}=\int \frac{sinxdx}{si{n}^{2}x(1+2cosx)}=\int \frac{sinxdx}{(1-cosx)(1+cosx)(1+2cosx)}$
Now differential coefficient of cos x is -sin x which is given in numerator and hence we make the substitution $cosx=t\Rightarrow -sinxdx=dt$
$\therefore I=-\int \frac{dt}{(1-t)(1+t)(1+2t)}$
We split the integrand into partial fractions
$\therefore I=-\int [\frac{1}{6(1-t)}-\frac{1}{2(1+t)}+\frac{4}{3(1+2t)}]dt$ etc.