Question

Find$\int \frac{secx}{1+cosecx}dx.$

Answer 1

Let $I=\int \frac{secx}{1+cosecx}dx=\int \frac{sinx}{cosx(1+sinx)}dx=\int \frac{sinxcosx}{(1-sinx)(1+sinx{)}^2}dx$
Put $sinx=t\Rightarrow cosxdx=dt\therefore I=\int \frac{t}{(1-t)(1+t{)}^2}dt$
Consider $\frac{t}{(1-t)(1+t{)}^2}=\frac{A}{1+t}+\frac{B}{(1+t{)}^2}+\frac{C}{1-t}\Rightarrow t=A(1-t^2)+B(1-t)+C(1+t{)}^2$
On comparing the coefficients of like terms on both sides, we get: $A=\frac{1}{4},B=-\frac{1}{2},C=\frac{1}{4}.$
So, $I=\int (\frac{1}{4}\times \frac{1}{1+t}-\frac{1}{2}\times \frac{1}{(1+t{)}^2}\times \frac{1}{4}\times \frac{1}{1-t})dt=\frac{1}{4}log|1+t|+\frac{1}{2+2t}-\frac{1}{4}log|1-t|+c$
$\therefore I=\frac{1}{4}log|\frac{1+sinx}{1-sinx}|+\frac{1}{2+2sinx}+c.$