Question

Integrate the function $ \cot x\log \left( \sin x \right) $ .

Answer 1

Hint: For this problem, we will use the substitution method. According to this method, we will take the substitution of the complex function in the given function and find the derivative of the taken substitution. Here, we will take $ \log \left( \sin x \right) $ as t and proceed. Then we will substitute both the values in the given function and apply the integration formulas to get the result.

Complete step by step answer:
Given function, $ \cot x\log \left( \sin x \right) $ .
Let $ f\left( x \right)=\cot x\log \left( \sin x \right) $ .
Integration of the above function is $ \int{f\left( x \right)}=\int{\cot x\log \left( \sin x \right)}dx $
In the above integration we have the complex function $ \log \left( \sin x \right) $ . So, taking the substitution $ \log \left( \sin x \right)=t...\left( \text{i} \right) $ .
Differentiating the above value with respect to $ x $ , we will get
 $ \dfrac{d}{dx}\left[ \log \left( \sin x \right) \right]=\dfrac{dt}{dx} $
We know that $ \dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right) $ . In the above equation the value $ f\left( g\left( x \right) \right)=\log \left( \sin x \right) $ i.e., $ f\left( x \right)=\log x $ , $ g\left( x \right)=\sin x $ . Now the derivatives of the both functions are given by $ {{f}^{'}}\left( x \right)=\dfrac{1}{x} $ and $ {{g}^{'}}\left( x \right)=\cos x $ .
 $ \therefore $ The value of $ \dfrac{dt}{dx} $ can be written as
 $ \begin{align}
  & \dfrac{d}{dx}\left[ \log \left( \sin x \right) \right]=\dfrac{dt}{dx} \\
 & \Rightarrow \dfrac{1}{\sin x}.\cos x=\dfrac{dt}{dx} \\
\end{align} $
We know that $ \dfrac{\cos x}{\sin x}=\cot x $ , then we will get
 $ \begin{align}
  & \dfrac{dt}{dx}=\cot x \\
 & \Rightarrow dt=\cot x.dx...\left( \text{ii} \right) \\
\end{align} $
From equation $ \left( \text{i} \right) $ and $ \left( \text{ii} \right) $ , the integration of the given function is written as
 $ \begin{align}
  & \int{f\left( x \right)}dx=\int{\cot x\log \left( \sin x \right)dx} \\
 & \Rightarrow \int{f\left( x \right)dx}=\int{\left( \log \left( \sin x \right) \right)\left( \cot xdx \right)} \\
 & \Rightarrow \int{f\left( x \right)}dx=\int{tdt} \\
\end{align} $
We have the integration formula $ \int{xdx}=\dfrac{{{x}^{2}}}{2}+C $ , then we will get
 $ \int{f\left( x \right)}dx=\dfrac{{{t}^{2}}}{2}+C $
From equation $ \left( \text{i} \right) $ substituting $ t=\log \left( \sin x \right) $ in the above equation, then we will get
 $ \int{f\left( x \right)}dx=\dfrac{{{\left( \log \left( \sin x \right) \right)}^{2}}}{2}+C $
 $ \therefore $ Integration value of the given function $ \cot x\log \left( \sin x \right) $ is $ \dfrac{{{\left( \log \left( \sin x \right) \right)}^{2}}}{2}+C $ .

Note:
We have a similar type of problem i.e., integrate the function $ f\left( x \right)=\dfrac{\cot x}{\log \left( \sin x \right)} $ . From this problem also we will use the same substitutions, then we will get
 $ \begin{align}
  & \int{f\left( x \right)dx}=\int{\dfrac{\cot x}{\log \left( \sin x \right)}dx} \\
 & \Rightarrow \int{f\left( x \right)dx}=\int{\dfrac{1}{\log \left( \sin x \right)}\left( \cot xdx \right)} \\
 & \Rightarrow \int{f\left( x \right)dx}=\int{\dfrac{1}{t}dt} \\
\end{align} $
We have the formula $ \int{\dfrac{1}{x}dx}=\log x+C $ , then we will get
 $ \begin{align}
  & \int{f\left( x \right)dx}=\log \left| t \right|+C \\
 & \Rightarrow \int{f\left( x \right)dx}=\log \left| \log \left( \sin x \right) \right|+C \\
\end{align} $
The small change in the operations in the given function results in a different integration value. In integration we have used the term $ C $ which is integration constant and the value of this constant calculated when the range of the integral or the limits of the integral are given.