Question

How do you find the indefinite integral of $\int{\dfrac{1}{\csc x}dx}$?

Answer 1

Hint: We first simplify the given expression of $\dfrac{1}{\csc x}$ to get $\dfrac{1}{\csc x}=\sin x$. We take the function $\cos x$ and find its differential form as $d\left( \cos x \right)=-\sin xdx$. The integration gives us the solution of $\int{\dfrac{1}{\csc x}dx}$ as $-\cos x+c$.


Complete step by step answer:
We first simplify the given expression where we get $\dfrac{1}{\csc x}=\sin x$.
We now try to find the differentiated form of $\cos x$.
We get $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$.
We get the differential form of $d\left( \cos x \right)=-\sin xdx$.
Changing the sign, we get
$\begin{align}
  & d\left( \cos x \right)=-\sin xdx \\
 & \Rightarrow -d\left( \cos x \right)=\sin xdx \\
 & \Rightarrow d\left( -\cos x \right)=\sin xdx \\
\end{align}$
Now we take the integration to get
$\begin{align}
  & \int{\sin xdx}=\int{d\left( -\cos x \right)}+c \\
 & \Rightarrow \int{\sin xdx}=-\cos x+c \\
\end{align}$
Therefore, the solution for $\int{\dfrac{1}{\csc x}dx}$ is $-\cos x+c$.
Note: We can also solve the integration using the base change for ratio $z=\cos x$. In that case the change in the variable goes as $dz=-\sin xdx$.
$\begin{align}
  & \int{\sin xdx} \\
 & =-\int{dz} \\
 & =-z+c \\
 & =-\cos x+c \\
\end{align}$