Question

Evaluate the following indefinite integral: $\int{\dfrac{\left( 1+\cot x \right)}{\left( x+\log \left( \sin x \right) \right)}dx}$?

Answer 1

Hint: We start solving the problem by assigning variable I to the given indefinite integral. We then assume $x+\log \left( \sin x \right)=t$ and apply a differential on both sides of this. We then make use of the results $d\left( a+b \right)=da+db$, $d\left( \log \left( f \right) \right)=\dfrac{1}{f}df$, $d\left( \sin x \right)=\cos xdx$, $d\left( \sin x \right)=\cos xdx$ to get the relation between $dx$ and $dt$. We then substitute $x+\log \left( \sin x \right)=t$ and relation of dx and dt in I, we then make use of the result $\int{\dfrac{1}{x}dx}=\log x+c$ and substitute $x+\log \left( \sin x \right)=t$ to get the required result.

Complete step-by-step answer:
According to the problem, we are asked to evaluate the given indefinite integral $\int{\dfrac{\left( 1+\cot x \right)}{\left( x+\log \left( \sin x \right) \right)}dx}$.
Let us assume $I=\int{\dfrac{\left( 1+\cot x \right)}{\left( x+\log \left( \sin x \right) \right)}dx}$ ---(1).
Let us assume $x+\log \left( \sin x \right)=t$ ---(2). Now, let us apply the differential on both sides.
We get $d\left( x+\log \left( \sin x \right) \right)=dt$ ---(3).
We know that $d\left( a+b \right)=da+db$. Let us apply this result in equation (3).
$\Rightarrow dx+d\left( \log \left( \sin x \right) \right)=dt$ ---(4).
We know that $d\left( \log \left( f \right) \right)=\dfrac{1}{f}df$. Let us apply this result in equation (4)
$\Rightarrow dx+\dfrac{1}{\sin x}d\left( \sin x \right)=dt$ ---(5).
We know that $d\left( \sin x \right)=\cos xdx$. Let us apply this result in equation (5).
$\Rightarrow dx+\dfrac{\cos x}{\sin x}dx=dt$ ---(6).
We know that $\dfrac{\cos x}{\sin x}=\cot x$. Let us apply this result in equation (6)
$\Rightarrow dx+\cot xdx=dt$.
$\Rightarrow \left( 1+\cot x \right)dx=dt$ ---(7).
Let us substitute equation (2) and (7) in equation (1).
So, we get $I=\int{\dfrac{1}{t}dt}$ ---(8).
We know that $\int{\dfrac{1}{x}dx}=\log x+c$. Let us apply this result in equation (8)
$\Rightarrow I=\log t+C$ ---(9).
From equation (2), we have $x+\log \left( \sin x \right)=t$. Let us substitute this in equation (9).
$\Rightarrow I=\log \left( x+\log \left( \sin x \right) \right)+C$, where C is a constant of integration.
So, we have found the result of indefinite integral as $\log \left( x+\log \left( \sin x \right) \right)+C$.
∴ $\int{\dfrac{\left( 1+\cot x \right)}{\left( x+\log \left( \sin x \right) \right)}dx}=\log \left( x+\log \left( \sin x \right) \right)+C$.

Note: We can see that the given problems contain huge amounts of calculations, so we need to perform each step carefully in order to avoid calculation mistakes and confusion. We should not confuse differential with differentiation as the results will vary a lot. We should not forget to add constant integration as it is the major mistake done by students. Similarly, we can expect problems to solve the definite integral \[\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\left( 1+\cot x \right)}{\left( x+\log \left( \sin x \right) \right)}dx}\].