Question

The integral $\int{\dfrac{\cot x}{\log \left( \sin x \right)}}dx$ equals:
(a) log (log(sin x)) + c
(b) log (log(sec x)) + c
(c) log (log(cot x)) + c
(d) log (log(cosec x)) + c

Answer 1

Hint: Take the denominator in the given expression as t. The denominator is log (sin x) equate it to t then take the derivative on both the sides. Then replace all the x expressions in the integral to t then the integral will reduce into t variable and after integration replace the t variable with x.

Complete step-by-step answer:
In the expression below, let us assume log (sin x) as t:
$\int{\dfrac{\cot x}{\log \left( \sin x \right)}}dx$
log (sin x) = t
Now, differentiating both the sides will give:
$\dfrac{1}{\sin x}\left( \cos x \right)dx=dt$
In the above expression, we are differentiating the left hand side by chain rule with respect to x.
We first assume sin x as u then the expression will be log (u) then differentiation is done as follows:
log (u) =t
$\dfrac{1}{u}du=dt$……………..Eq. (1)
We know that the derivative of log x with respect to x is$\dfrac{1}{x}$. Now, we are converting “u” variable to x as follows:
u = sin x
Differentiating both the sides we get,
du = cos x dx
Substituting the above value of du and u in the eq. (1) we get,
$\begin{align}
  & \dfrac{1}{\sin x}\cos xdx=dt \\
 & \Rightarrow \cot xdx=dt \\
\end{align}$
Now, substituting the values of cot x dx as dt and log (sin x) as t which we have solved above in this expression $\int{\dfrac{\cot x}{\log \left( \sin x \right)}}dx$ we get,
$\int{\dfrac{dt}{t}}$
We know that, integration of $\dfrac{1}{t}$ with respect to t gives log t so the above expression will resolve to:
log (t) + c
Now, substitute the value of t in the above expression we get,
log (log (sin x)) + c
From the above solution, the value of$\int{\dfrac{\cot x}{\log \left( \sin x \right)}}dx$ is log (log (sin x)) + c.
Hence, the correct option is (a).

Note: There is a trick of solving rational types integral by taking denominator as t and the derivative of denominator we can see in numerator of the integral then we replace the numerator by some multiple of dt and it will make the integration easier as we have shown in the above solution.