Question

Solve \[\int {\dfrac{{\sin x}}{{\left( {\sin x - \cos x} \right)}}} dx\]

Answer 1

Hint: Firstly we will assume that the given integral be \[I\] and then multiply and divide the integral by \[2\] and split the term \[2\sin x\] in the terms of \[\sin x\] and \[\cos x\]. After that find the integral by assuming the simplified integral as \[t\] then simplify it by finding the differential and by putting the value of obtained differential in the integral to obtain the given result.

Complete step by step answer:
The calculation of an integral is known as integration. Integrals are used in mathematics to calculate a variety of useful quantities such as areas, volumes, displacement, and so on. When we talk about integrals, we usually mean definite integrals. Antiderivatives are computed using indefinite integrals. Hence apart from differentiation, integration is one of the two major calculus subjects in mathematics which measure the rate of change of any function with respect to its variables. Integration is a way of finding the whole by adding or summing the parts. It's a reversal of differentiation, in which we break down functions into pieces. This approach is used to calculate the sum on a large scale. Calculating little addition problems is a simple operation that can be done manually or with the aid of calculators. Integration methods, on the other hand, are utilised for large addition problems where the bounds could reach infinite. Calculus is divided into two parts: integration and differentiation. These topics have a very high idea level. As a result, it is first exposed to us in high school and later in engineering or further education. The area of the region encompassed by the graph of functions is defined and calculated using integration. As a result, we have two types of integrals: indefinite and definite.
Now according to the question:
We have to solve the integral \[\int {\dfrac{{\sin x}}{{\left( {\sin x - \cos x} \right)}}} dx\]
Let the integral be \[I\] such that:
\[ \Rightarrow I = \int {\dfrac{{\sin x}}{{\left( {\sin x - \cos x} \right)}}} dx\]
Multiply and divide the integral by \[2\] we get:
\[ \Rightarrow I = \int {\dfrac{{2\sin x}}{{2\left( {\sin x - \cos x} \right)}}} dx\]
\[ \Rightarrow I = \dfrac{1}{2}\int {\dfrac{{2\sin x}}{{\left( {\sin x - \cos x} \right)}}} dx\]
Split \[2\sin x\] as \[\left( {\sin x - \cos x} \right) + \left( {\sin x + \cos x} \right)\]
\[ \Rightarrow I = \dfrac{1}{2}\int {\dfrac{{\left( {\sin x - \cos x} \right) + \left( {\sin x + \cos x} \right)}}{{\left( {\sin x - \cos x} \right)}}} dx\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\int {\dfrac{{\left( {\sin x - \cos x} \right)}}{{\left( {\sin x - \cos x} \right)}}} dx + \int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left( {\sin x - \cos x} \right)}}dx} } \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\int {dx} + \int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left( {\sin x - \cos x} \right)}}dx} } \right]\]
Put \[\left( {\sin x - \cos x} \right) = t\]
Hence the differential will be written as:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x - \cos x} \right) = \dfrac{d}{{dx}}t\]
\[ \Rightarrow \left( {\cos x - \left( { - \sin x} \right)} \right) = \dfrac{{dt}}{{dx}}\]
\[ \Rightarrow \left( {\cos x + \sin x} \right) = \dfrac{{dt}}{{dx}}\]
\[ \Rightarrow \left( {\cos x + \sin x} \right)dx = dt\]
Now put all the values in the integral we get:
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\int {dx} + \int {\dfrac{1}{t}dt} } \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {x + \ln t} \right] + c\]
Put the value of \[t\] that is \[\left( {\sin x - \cos x} \right)\] we get:
\[ \Rightarrow I = \dfrac{1}{2}\left[ {x + \ln \left( {\sin x - \cos x} \right)} \right] + c\]
\[ \Rightarrow I = \dfrac{x}{2} + \dfrac{{\ln \left( {\sin x - \cos x} \right)}}{2} + c\]
Where \[c\] is any constant.

Note:
In this problem, we have re-written the actual terms for our convenience like \[\int {\dfrac{{\sin x}}{{\left( {\sin x - \cos x} \right)}}} dx\] as \[\int {\dfrac{{2\sin x}}{{2\left( {\sin x - \cos x} \right)}}} dx\] , \[2\sin x\] as \[\left( {\sin x - \cos x} \right) + \left( {\sin x + \cos x} \right)\]these are done to make our solution progress easier. Also, while using substitution method remember that we need to re-substitute it to get the original terms.