Question

Integrate \[\dfrac{1}{{1 + \cot x}}\] using the substitution method.

Answer 1

Hint: Here we have indefinite integral, before applying integration we need to simplify the integrand using the definition of cotangent that is \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]. After we simplify by taking LCM in the denominator of the integrand and we split the terms. Then we can apply the substitution method and we need to make sure that the final answer should be in terms of ‘x’ only.

Complete step by step answer:
Consider
\[I = \int {\left( {\dfrac{1}{{1 + \cot x}}} \right)dx} \]
Now using the definition of cotangent we have \[\cot x = \dfrac{{\cos x}}{{\sin x}}\], then above becomes
\[I = \int {\left( {\dfrac{1}{{1 + \left( {\dfrac{{\cos x}}{{\sin x}}} \right)}}} \right)dx} \]
Taking LCM and simplifying we have,
\[I = \int {\left( {\dfrac{1}{{\left( {\dfrac{{\sin x + \cos x}}{{\sin x}}} \right)}}} \right)dx} \]
\[\Rightarrow I = \int {\left( {\dfrac{{\sin x}}{{\sin x + \cos x}}} \right)dx} \]
Now to simplify further multiply and divide by 2 in the integrand,
\[I = \int {\dfrac{1}{2}\left( {\dfrac{{2\sin x}}{{\sin x + \cos x}}} \right)dx} \]
\[\Rightarrow I = \dfrac{1}{2}\int {\left( {\dfrac{{2\sin x}}{{\sin x + \cos x}}} \right)dx} \]

We can write \[2\sin x = \sin x + \sin x\]
\[I = \dfrac{1}{2}\int {\left( {\dfrac{{\sin x + \sin x}}{{\sin x + \cos x}}} \right)dx} \]
Now add and subtract \[\cos x\] on the numerator of the integrand,
\[I = \dfrac{1}{2}\int {\left( {\dfrac{{\sin x + + \cos x + \sin x - \cos x}}{{\sin x + \cos x}}} \right)dx} \]
Grouping we have,
\[I = \dfrac{1}{2}\int {\left( {\dfrac{{\left( {\sin x + \cos x} \right) + \left( {\sin x - \cos x} \right)}}{{\sin x + \cos x}}} \right)dx} \]
Now splitting,
\[I = \dfrac{1}{2}\int {\left( {\dfrac{{\sin x + \cos x}}{{\sin x + \cos x}}} \right)dx} + \dfrac{1}{2}\int {\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)dx} \]

Cancelling we have
\[I = \dfrac{1}{2}\int {dx} + \dfrac{1}{2}\int {\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)dx} \]
Now for the second integral we use the substitution method. That is we substitute
\[\sin x + \cos x = t\]
Now differentiating with respect to x
\[\left( {\cos x - \sin x} \right)dx = dt\]. Then above integration becomes
\[\Rightarrow I = \dfrac{1}{2}\int {dx} + \dfrac{1}{2}\int {\dfrac{1}{t}dt} \]
\[\Rightarrow I = \dfrac{1}{2}x + \dfrac{1}{2}\log |t| + C\]
Where C is the integration constant.
But in the end we need answers in terms of x only. We substituted \[t = \sin x + \cos x\]
\[\therefore I = \dfrac{1}{2}x + \dfrac{1}{2}\log |\sin x + \cos x| + C\]

Hence integration of \[\dfrac{1}{{1 + \cot x}}\] is \[\dfrac{1}{2}x + \dfrac{1}{2}\log |\sin x + \cos x| + C\].

Note: In definite integral we will not have integration constant C due to the presence of upper and lower limits. Also integration can be used to find the area between the curves, volumes, central points and many useful. We also know the derivative of a constant is zero but not for integration.