Answer
Verified
398.4k+ 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
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE