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Question

Question: Find the integration of \( \int {\cot x \cdot \ln (\sin x) \cdot dx} \) ....

Find the integration of cotxln(sinx)dx\int {\cot x \cdot \ln (\sin x) \cdot dx} .

Explanation

Solution

Hint : To solve this problem, we need to use the method of integration by substitution. In this method, we can find an integral but only when it can be set up in a certain way. The most important and first step to apply this method is to be able to write the integral in the form of: f(g(x))g(x)dx\int {f(g(x)) \cdot g'(x) \cdot dx}

Complete step-by-step answer :
Here, we need to solve I=cotxln(sinx)dxI = \int {\cot x \cdot \ln (\sin x) \cdot dx}
To apply the method of integration by substitution, we will first set up the integral in the form of f(g(x))g(x)dx\int {f(g(x)) \cdot g'(x) \cdot dx}
Let us take g(x)=ln(sinx)g(x) = \ln (\sin x) .
Now, to differentiate this with respect to xx , we need to apply the chain rule.
We know that the differentiation of lnx\ln x is 1x\dfrac{1}{x} and the differentiation of sinx\sin x is cosx\cos x .
g(x)=1sinxcosx=cosxsinx=cotx\Rightarrow g'(x) = \dfrac{1}{{\sin x}} \cdot \cos x = \dfrac{{\cos x}}{{\sin x}} = \cot x
Therefore, we can set up the given integral in the desired form to apply the substitution method of integration.
Now, let us take u=g(x)u = g(x)
We have already taken g(x)=ln(sinx)g(x) = \ln (\sin x)
u=ln(sinx)\Rightarrow u = \ln (\sin x)
Differentiate with respect to xx
dudx=1sinxcosx dudx=cotx du=cotxdx   \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sin x}} \cdot \cos x \\\ \Rightarrow \dfrac{{du}}{{dx}} = \cot x \\\ \Rightarrow du = \cot xdx \;
Therefore, if we write our integral in terms of uu, we get
I=uduI = \int {udu}
We know that the integration of xn{x^n} is xn+1n+1+c\dfrac{{{x^{n + 1}}}}{{n + 1}} + c, therefore integration of uu becomes u22+c\dfrac{{{u^2}}}{2} + c , where cc is a constant.
I=u22+c\Rightarrow I = \dfrac{{{u^2}}}{2} + c
We have taken u=ln(sinx)u = \ln (\sin x)
I=12(lnsinx)2+c\Rightarrow I = \dfrac{1}{2}{\left( {\ln \sin x} \right)^2} + c , where cc is a constant.
Thus our final answer is: cotxln(sinx)dx=12(lnsinx)2+c\int {\cot x \cdot \ln (\sin x) \cdot dx} = \dfrac{1}{2}{\left( {\ln \sin x} \right)^2} + c , where cc is a constant.
So, the correct answer is “12(lnsinx)2+c\dfrac{1}{2}{\left( {\ln \sin x} \right)^2} + c ”.

Note : Here, we have applied the method of integration by substitution to solve the problem. There are three important steps to remember while applying this method.
The first step to begin the solution is to set up the integral in this form: f(g(x))g(x)dx\int {f(g(x)) \cdot g'(x) \cdot dx}
The second step is to make u=g(x)u = g(x) and then integrate f(u)du\int {f(u)du}
The final step is to reinsert g(x)g(x) in place of uu .