Question

How do find the indefinite integral of $\int {(\cot x)\dfrac{{dx}}{{1 + \cos x}}dx} $

Answer 1

Hint:Convert cot(x) in terms of $\sin (x)$ and $\cos (x)$, then continue integration.In this problem, we first break the down the function in two simpler form, then we break the function into two half and solve them separately then if there is a product-based denominator which should be integrated, we use Laplace transformation, to get solve the equation and make into simpler form and later after solving the two parts separately, we combine them and get the final answer. This process is also called integration by parts.

Complete step by step answer:
Here, as discussed, convert the \[cot(x)\]in terms of $\sin (x)$ and $\cos (x)$ and integrate it.
\[I = \int {\dfrac{{\cot x}}{{1 + \cos x}}dx} \\
\Rightarrow I = \int {\dfrac{{\cos x}}{{\sin x(1 + \cos x)}}dx} \\
\Rightarrow I = \int {\dfrac{{(1 + \cos x) - 1}}{{\sin x(1 + \cos x)}}dx} \\
\Rightarrow I = \int {\dfrac{1}{{\sin x}}dx - \int {\dfrac{1}{{\sin x(1 + \cos x)}}} dx} \\
\Rightarrow I = \int {\csc xdx - \int {\dfrac{{\sin x}}{{{{\sin }^2}x(1 + \cos x)}}} } \\
\Rightarrow I= \ln \left| {\csc x - \cot x} \right| - \int {\dfrac{{\sin x}}{{(1 - {{\cos }^2}x)(1 + \cos x)}}} \\
\Rightarrow I = \ln \left| {\csc x - \cot x} \right| - {I_1} \\ \]..........................Equation 1
Where ${I_1} = \int {\dfrac{{\sin x}}{{(1 - \cos x){{(1 + \cos x)}^2}}}} $
Let \[cos{\text{(}}x)\]be equal to u and \[sin(x)\]dx be equal to -du.
So, we get the new form as,
${I_1} = \int {\dfrac{{ - 1}}{{(1 - u){{(1 + u)}^2}}}du} $
Now, we are going to split the denominator into three parts and use Laplace transformation
$\dfrac{{ - u}}{{(1 - u){{(1 + u)}^2}}} = \dfrac{A}{{1 - u}} + \dfrac{B}{{1 + u}} + \dfrac{C}{{{{(1 + u)}^2}}} \\
\Rightarrow - 1 = A{(1 + u)^2} + B((1 - u)(1 + u)) + C(1 - u) \\ $
When u=1, we get
$A{(1 + 1)^2} = - 1 \\
\Rightarrow A = \dfrac{{ - 1}}{4} \\ $
When u= -1, we get
$C(1 + 1) = - 1 \\
\Rightarrow C = \dfrac{{ - 1}}{2} \\ $
When u=0, we get
$A(1) + B(1) + C(1) = - 1 \\
\Rightarrow A + B + C = - 1 \\ $
In the above obtained equation, we substitute the values of A and C to get the value of B
$\left( {\dfrac{{ - 1}}{4}} \right) + B + \left( {\dfrac{{ - 1}}{2}} \right) = - 1 \\
\Rightarrow B = - \dfrac{1}{4} \\ $
Now that we have obtained all the values needed in the Laplace transformation, we substitute them, as follows
${I_1} = \int {\dfrac{{ - \dfrac{1}{4}}}{{1 - u}}du + \int {\dfrac{{ - \dfrac{1}{4}}}{{1 + u}}du + \int {\dfrac{{ - \dfrac{1}{2}}}{{{{(1 + u)}^2}}}du} } } \\
\Rightarrow{I_1} = \dfrac{1}{4}\int {\dfrac{1}{{u - 1}}du - \dfrac{1}{4}\int {\dfrac{1}{{1 + u}}du - \dfrac{1}{2}\int {{{(1 + u)}^{ - 2}}du} } } \\
\Rightarrow{I_1}= \dfrac{1}{4}\ln \left| {u - 1} \right| - \dfrac{1}{4}\ln \left| {u + 1} \right| - \dfrac{1}{2}\left( {\dfrac{{{{(1 + u)}^{ - 1}}}}{{ - 1}}} \right) + c \\$
Here in the next step, we use the log properties and proceed. As previously we had considered u in the place of \[cos(x)\], now we substitute it back,
${I_1} = \dfrac{1}{4}\ln \left| {\dfrac{{cox - 1}}{{\cos x + 1}} + \dfrac{1}{{2(1 + \cos x)}} + c} \right|$
As, we have a simplified and integrated value of ${I_1}$ , now we are substitute it back in the equation 1
$\therefore I = \ln \left| {\csc x - \cot x} \right| - \dfrac{1}{4}\ln \left| {\dfrac{{\cos x - 1}}{{\cos x + 1}}} \right| + \dfrac{1}{{2(1 + \cos x)}} + c$
Where $c$ is the integration constant.

Hence, the indefinite integral of $\int {(\cot x)\dfrac{{dx}}{{1 + \cos x}}dx}$ is $\ln \left| {\csc x - \cot x} \right| - \dfrac{1}{4}\ln \left| {\dfrac{{\cos x - 1}}{{\cos x + 1}}} \right| + \dfrac{1}{{2(1 + \cos x)}} + c$.

Note:The problems like these which are complex and need to be broken down into parts, we should always follow integration by parts, as it makes it easy for solving the problem, as we substitute them, break them and accordingly differentiate and integrate to get the final solution by combining them back.