Question

Evaluate : $\int \frac{\sec x}{1+\text{cosec}x}dx$

Answer 1

$\int \frac{\sec x}{1+\text{cosec}x}dx=\int \frac{1}{\cos x\left(\begin{array}{c}1+\frac{1}{\sin x}\end{array}\right)}dx$

$=\int \frac{\sin x}{\cos x(\sin x+1)}dx$

$=\int \frac{\sin x\cos x}{{\cos }^{2}x(1+\sin x)}dx$

$=\int \frac{\sin x\cos x}{(1-{\sin }^{2}x)(1+\sin x)}dx$

$=\int \frac{\sin x\cos x}{(1-\sin x)(1+\sin x)(1+\sin x)}dx$

$=\int \frac{\sin x\cos x}{(1-\sin x)(1+\sin x{)}^2}dx$

Now, let $\sin x=t$, on differentiating, we get
$\cos xdx=dt$
$\therefore \int \frac{\sec x}{1+\text{cosec}x}dx=\int \frac{t}{(1+t)(1+t{)}^2}dt$

Let $\frac{t}{(1-t)(1+t{)}^2}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{(1+t{)}^2}$

$\Rightarrow t=A(1+t{)}^2+B(1-t\left)\right(1+t)+C(1-t)$ ...$\left(i\right)$

On putting $t=1$ and $t=-1$, we get
$1=A\cdot 4$ and $-1=C\cdot 2$
$\Rightarrow A=\frac{1}{4}$ and $C=-\frac{1}{2}$

On equating coefficient of $t^2$ on both sides of $\left(i\right)$, we get
$0=A-B$
$\Rightarrow B=A=\frac{1}{4}$
$\therefore \int \frac{\sec x}{1+\text{cosec}x}dx=\frac{1}{4}\int \frac{1}{1-t}dt+\frac{1}{4}\int \frac{dt}{1+t}-\frac{1}{2}\int \frac{dt}{(1+t{)}^2}$

$=\frac{1}{4}\frac{\log |1-t|}{-1}+\frac{1}{4}\log |1+t|-\frac{1}{2}\frac{(1+t{)}^{-1}}{-1}+c$

$=\frac{1}{4}\log \left|\begin{array}{c}\frac{1+t}{1-t}\end{array}\right|+\frac{1}{2(1+t)}+c$

$=\frac{1}{4}\log \left|\begin{array}{c}\frac{1+\sin x}{1-\sin x}\end{array}\right|+\frac{1}{2(1+\sin x)}+c$.