Question

Find the value of integral \[\int{\dfrac{\cot x}{\log \left( \sin x \right)}dx}\]
(a) \[\log \left( \log \left( \sin x \right) \right)+c\]
(b) \[\log \left( \log \left( \sec x \right) \right)+c\]
(c) \[\log \left( \log \left( \cot x \right) \right)+c\]
(d) \[\log \left( \log \left( \csc x \right) \right)+c\]

Answer 1

Hint: We solve this problem by using the substituting method.
We assume some function of the given integral as some other variable so that we can convert the given integral of \[x\] into the integral of the assumed variable which gives the standard form.
We differentiate the assumed function to get the value of \[dx\]
We use the chain rule of differentiation that is
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)\]
We have some standard results of differentiation as
\[\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\]
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]

Complete step by step answer:
We are asked to find the integral value of \[\int{\dfrac{\cot x}{\log \left( \sin x \right)}dx}\]
Let us assume that the given integral as
\[\Rightarrow I=\int{\dfrac{\cot x}{\log \left( \sin x \right)}dx}\]
Let us solve this problem by using the substitution method.
Let us assume that the denominator of given integral as
\[\Rightarrow \log \left( \sin x \right)=t\]
Now, let us differentiate the above equation on both sides with respect to \[x\] then we get
\[\Rightarrow \dfrac{d}{dx}\left( \log \left( \sin x \right) \right)=\dfrac{dt}{dx}\]
We know that the chain rule of differentiation that is
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)\]
We also know that the standard results of differentiation as
\[\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\]
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
By using this chain rule and standard results of differentiation to above equation then we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\sin x}\dfrac{d}{dx}\left( \sin x \right)=\dfrac{dt}{dx} \\
 & \Rightarrow \dfrac{1}{\sin x}\left( \cos x \right)=\dfrac{dt}{dx} \\
 & \Rightarrow \cot x.dx=dt \\
\end{align}\]
Here, we can see that we got the value of \[dx\]
Now, by substituting the required functions in given integral then we get
\[\Rightarrow I=\int{\dfrac{dt}{t}}\]
We know that the standard result of integration that is
\[\int{\dfrac{dx}{x}}=\log x+c\]
By using this result in above equation we get
\[\begin{align}
  & \Rightarrow I=\log t+c \\
 & \Rightarrow I=\log \left( \log \left( \sin x \right) \right)+c \\
\end{align}\]
Therefore, we can conclude that the value of given integral as
\[\therefore \int{\dfrac{\cot x}{\log \left( \sin x \right)}dx}=\log \left( \log \left( \sin x \right) \right)+c\]
So, option a) is the correct answer.


Note:
 We need to note that in the substitution method we assume some part of the given integral as some other variable but not the whole function in the integral.
We are using this substitution method so that we can convert the given complex form of the integral function to the standard form of integration.
If we take the whole function as some other variable then again it comes back to a complex form which will be of no use. So, we need to assume in such a way that we get an easy and standard form of integration.