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(A) $2\left( \sin x+x\cos \theta \right)+C$

(B) $2\left( \sin x-x\cos \theta \right)+C$

(C) $2\left( \sin x+2x\cos \theta \right)+C$

(D) $2\left( \sin x-2x\cos \theta \right)+C$

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Hint: Use trigonometric identity given as $\cos 2\theta =2{{\cos }^{2}}\theta -1$. Please note that $\cos 2\theta $ and $\cos \theta $are constants.

Here, we have integration given;

$\int{\dfrac{\cos 2x-\cos 2\theta }{\cos x-\cos \theta }}dx$……………………(1)

As, we can notice that ‘x’ is acting as a variable and $\theta $ is a constant as integration is given with respect to dx.

We can use trigonometric identity $\cos 2\theta =2{{\cos }^{2}}\theta -1$ to simplify the integration given in equation (1).

Hence, replacing $\cos 2x\ \text{by }2{{\cos }^{2}}x-1\ \text{and }\cos 2\theta \text{ by }2{{\cos }^{2}}\theta -1$ by using the above mentioned trigonometric identity.

Hence, equation (1) or given integration can be written as;

\[\begin{align}

& \int{\dfrac{\left( 2{{\cos }^{2}}x-1 \right)-\left( 2{{\cos }^{2}}\theta -1 \right)}{\cos x-\cos \theta }}dx \\

& \int{\dfrac{2{{\cos }^{2}}x-1-2{{\cos }^{2}}\theta +1}{\cos x-\cos \theta }}dx \\

& \int{\dfrac{2\left( {{\cos }^{2}}x-{{\cos }^{2}}\theta \right)}{\cos x-\cos \theta }}dx..................\left( 2 \right) \\

\end{align}\]

Now, we have an algebraic identity given as;

${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ ………………….(3)

Now, we can use the above algebraic identity as mentioned in equation (3) with equation (2). We can replace \[{{\cos }^{2}}x-{{\cos }^{2}}\theta \text{ by }\left( \cos x-\cos \theta \right)\left( \cos x+\cos \theta \right)\].

Hence, equation (2) can be written as;

\[\int{\dfrac{2\left( \cos x-\cos \theta \right)\left( \cos x+\cos \theta \right)}{\left( \cos x-\cos \theta \right)}}dx\]

Now, cancelling same times from numerator and denominator, we get;

\[2\int{\left( \cos x-\cos \theta \right)dx}\] …………………….(4)

Now, we know that;

$\int{\left( f\left( x \right)+g\left( x \right) \right)dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx}}}$

Hence, we can write equation (4), as

\[2\int{\cos xdx}+2\int{\cos \theta dx}\]

We know integration of cos x is sin x and $\cos \theta $ is acting as a constant with respect to ‘x’. Hence, we can take it out of integration in the above equation, we get;

$\begin{align}

& 2\sin x+2\cos \theta \int{1dx} \\

& or \\

& 2\sin x+2\cos \theta \int{{{x}^{0}}dx} \\

\end{align}$

As we know $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$, Hence $\int{{{x}^{0}}dx=\dfrac{{{x}^{0+1}}}{0+1}}=x$

Therefore, above equation can be written as;

$\begin{align}

& 2\sin x+2\cos \theta \left( x \right)+C \\

& or \\

& 2\left( \sin x+x\cos \theta \right)+C \\

& Hence, \\

& \int{\dfrac{\cos 2x-\cos 2\theta }{\cos x-\cos \theta }}=2\left( \sin x+x\cos \theta \right)+C \\

\end{align}$

So, option (A) is the correct answer.

Note: One can apply the trigonometric identity of $\cos 2\theta \text{ as }1-2{{\cos }^{2}}\theta $ which will give the wrong solution. Hence, use $\cos 2\theta =2{{\cos }^{2}}\theta -1$ .

Another approach for this question would be that we can multiply by $\cos x+\cos \theta $ in numerator and denominator, we get;

\[\int{\dfrac{\left( \cos 2x-\cos 2\theta \right)\left( \cos x+\cos \theta \right)}{\left( \cos x-\cos \theta \right)\left( \cos x+\cos \theta \right)}}dx\]

Using relation $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, and solve accordingly.

Here, we have integration given;

$\int{\dfrac{\cos 2x-\cos 2\theta }{\cos x-\cos \theta }}dx$……………………(1)

As, we can notice that ‘x’ is acting as a variable and $\theta $ is a constant as integration is given with respect to dx.

We can use trigonometric identity $\cos 2\theta =2{{\cos }^{2}}\theta -1$ to simplify the integration given in equation (1).

Hence, replacing $\cos 2x\ \text{by }2{{\cos }^{2}}x-1\ \text{and }\cos 2\theta \text{ by }2{{\cos }^{2}}\theta -1$ by using the above mentioned trigonometric identity.

Hence, equation (1) or given integration can be written as;

\[\begin{align}

& \int{\dfrac{\left( 2{{\cos }^{2}}x-1 \right)-\left( 2{{\cos }^{2}}\theta -1 \right)}{\cos x-\cos \theta }}dx \\

& \int{\dfrac{2{{\cos }^{2}}x-1-2{{\cos }^{2}}\theta +1}{\cos x-\cos \theta }}dx \\

& \int{\dfrac{2\left( {{\cos }^{2}}x-{{\cos }^{2}}\theta \right)}{\cos x-\cos \theta }}dx..................\left( 2 \right) \\

\end{align}\]

Now, we have an algebraic identity given as;

${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ ………………….(3)

Now, we can use the above algebraic identity as mentioned in equation (3) with equation (2). We can replace \[{{\cos }^{2}}x-{{\cos }^{2}}\theta \text{ by }\left( \cos x-\cos \theta \right)\left( \cos x+\cos \theta \right)\].

Hence, equation (2) can be written as;

\[\int{\dfrac{2\left( \cos x-\cos \theta \right)\left( \cos x+\cos \theta \right)}{\left( \cos x-\cos \theta \right)}}dx\]

Now, cancelling same times from numerator and denominator, we get;

\[2\int{\left( \cos x-\cos \theta \right)dx}\] …………………….(4)

Now, we know that;

$\int{\left( f\left( x \right)+g\left( x \right) \right)dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx}}}$

Hence, we can write equation (4), as

\[2\int{\cos xdx}+2\int{\cos \theta dx}\]

We know integration of cos x is sin x and $\cos \theta $ is acting as a constant with respect to ‘x’. Hence, we can take it out of integration in the above equation, we get;

$\begin{align}

& 2\sin x+2\cos \theta \int{1dx} \\

& or \\

& 2\sin x+2\cos \theta \int{{{x}^{0}}dx} \\

\end{align}$

As we know $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$, Hence $\int{{{x}^{0}}dx=\dfrac{{{x}^{0+1}}}{0+1}}=x$

Therefore, above equation can be written as;

$\begin{align}

& 2\sin x+2\cos \theta \left( x \right)+C \\

& or \\

& 2\left( \sin x+x\cos \theta \right)+C \\

& Hence, \\

& \int{\dfrac{\cos 2x-\cos 2\theta }{\cos x-\cos \theta }}=2\left( \sin x+x\cos \theta \right)+C \\

\end{align}$

So, option (A) is the correct answer.

Note: One can apply the trigonometric identity of $\cos 2\theta \text{ as }1-2{{\cos }^{2}}\theta $ which will give the wrong solution. Hence, use $\cos 2\theta =2{{\cos }^{2}}\theta -1$ .

Another approach for this question would be that we can multiply by $\cos x+\cos \theta $ in numerator and denominator, we get;

\[\int{\dfrac{\left( \cos 2x-\cos 2\theta \right)\left( \cos x+\cos \theta \right)}{\left( \cos x-\cos \theta \right)\left( \cos x+\cos \theta \right)}}dx\]

Using relation $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, and solve accordingly.

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