
Find $\int{\dfrac{\sin x}{\sin \left( x-a \right)}dx}$.
Answer
496.5k+ views
Hint: We first break the numerator of the fraction $\dfrac{\sin x}{\sin \left( x-a \right)}$ as $\sin x=\sin \left\{ \left( x-a \right)+a \right\}$. We apply the identity formula of $\sin \left( m+n \right)=\sin m\cos n+\cos m\sin n$. We use the integral formula of $\int{\cot xdx}=\log \left| \sin x \right|$. We break the integration and find the solution.
Complete step by step solution:
To simplify the term $\dfrac{\sin x}{\sin \left( x-a \right)}$, we first form the numerator as $\sin x=\sin \left\{ \left( x-a \right)+a \right\}$.
So, $\dfrac{\sin x}{\sin \left( x-a \right)}=\dfrac{\sin \left\{ \left( x-a \right)+a \right\}}{\sin \left( x-a \right)}$.
Now we use the trigonometric associative form of $\sin \left( m+n \right)=\sin m\cos n+\cos m\sin n$.
Taking the variables as $m=\left( x-a \right),n=a$, we get $\sin x=\sin \left( x-a \right)\cos a+\cos \left( x-a \right)\sin a$.
The simplified form will be $\dfrac{\sin x}{\sin \left( x-a \right)}=\dfrac{\sin \left( x-a \right)\cos a+\cos \left( x-a \right)\sin a}{\sin \left( x-a \right)}=\cos a+\cot \left( x-a \right)\sin a$
In the given terms, $a$ is constant and $x$ is variable. Therefore, both $\cos a,\sin a$ are constant.
So, $\int{\dfrac{\sin x}{\sin \left( x-a \right)}dx}=\int{\left[ \cos a+\cot \left( x-a \right)\sin a \right]dx}$.
We break the addition and get $\int{\left[ \cos a+\cot \left( x-a \right)\sin a \right]dx}=\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)dx}$.
We take the differential form as $d\left( x-a \right)=dx$.
We also know that $\int{\cot xdx}=\log \left| \sin x \right|$.
Therefore, $\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)dx}=\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)d\left( x-a \right)}$.
$\begin{align}
& \int{\dfrac{\sin x}{\sin \left( x-a \right)}dx} \\
& =\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)d\left( x-a \right)} \\
& =x\cos a+\sin a\log \left| \sin \left( x-a \right) \right|+c \\
\end{align}$
Here $c$ is the integral constant.
Note:
We broke the numerator instead of the denominator as that helps in breaking the fraction into two parts, one of which is constant. We need to change the differential form as the main formula of $\int{\cot xdx}=\log \left| \sin x \right|$ is for variable $x$.
Complete step by step solution:
To simplify the term $\dfrac{\sin x}{\sin \left( x-a \right)}$, we first form the numerator as $\sin x=\sin \left\{ \left( x-a \right)+a \right\}$.
So, $\dfrac{\sin x}{\sin \left( x-a \right)}=\dfrac{\sin \left\{ \left( x-a \right)+a \right\}}{\sin \left( x-a \right)}$.
Now we use the trigonometric associative form of $\sin \left( m+n \right)=\sin m\cos n+\cos m\sin n$.
Taking the variables as $m=\left( x-a \right),n=a$, we get $\sin x=\sin \left( x-a \right)\cos a+\cos \left( x-a \right)\sin a$.
The simplified form will be $\dfrac{\sin x}{\sin \left( x-a \right)}=\dfrac{\sin \left( x-a \right)\cos a+\cos \left( x-a \right)\sin a}{\sin \left( x-a \right)}=\cos a+\cot \left( x-a \right)\sin a$
In the given terms, $a$ is constant and $x$ is variable. Therefore, both $\cos a,\sin a$ are constant.
So, $\int{\dfrac{\sin x}{\sin \left( x-a \right)}dx}=\int{\left[ \cos a+\cot \left( x-a \right)\sin a \right]dx}$.
We break the addition and get $\int{\left[ \cos a+\cot \left( x-a \right)\sin a \right]dx}=\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)dx}$.
We take the differential form as $d\left( x-a \right)=dx$.
We also know that $\int{\cot xdx}=\log \left| \sin x \right|$.
Therefore, $\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)dx}=\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)d\left( x-a \right)}$.
$\begin{align}
& \int{\dfrac{\sin x}{\sin \left( x-a \right)}dx} \\
& =\cos a\int{dx}+\sin a\int{\cot \left( x-a \right)d\left( x-a \right)} \\
& =x\cos a+\sin a\log \left| \sin \left( x-a \right) \right|+c \\
\end{align}$
Here $c$ is the integral constant.
Note:
We broke the numerator instead of the denominator as that helps in breaking the fraction into two parts, one of which is constant. We need to change the differential form as the main formula of $\int{\cot xdx}=\log \left| \sin x \right|$ is for variable $x$.
Recently Updated Pages
Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

When was the first election held in India a 194748 class 12 sst CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

