Find $\int{\dfrac{\sin x}{\sin \left( x-a \right)}dx}$.
Last updated date: 20th Mar 2023
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Answer
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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 $x$.
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