Question

\(\int \frac{\cos(2x) - \cos(2\alpha)}{\cos(x) - \cos(\alpha)} \, dx\) is equal to

• A. \(2(\sin(x) - x \cos(\alpha)) + c\)
• B. \(2(\sin(x) - 2x \cos(\alpha)) + c\)
• C. \(2(\sin(x) + x \cos(\alpha)) + c\)
• D. \(2(\sin(x) + 2x \cos(\alpha)) + c\)

Answer 1

The Correct Option is C

We are tasked with evaluating the integral:

\[ \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx \]

We can use the identity \(\cos 2\theta = 2 \cos^2 \theta - 1\) to rewrite \(\cos 2x\) as:

\[ \cos 2x = 2 \cos^2 x - 1 \]

Similarly, \(\cos 2\alpha\) can be written as:

\[ \cos 2\alpha = 2 \cos^2 \alpha - 1 \]

Substituting these expressions into the integrand, we have:

\[ \frac{2 \cos^2 x - 1 - 2 \cos^2 \alpha + 1}{\cos x - \cos \alpha} \]

Simplifying further:

\[ \frac{2 \cos^2 x - 2 \cos^2 \alpha}{\cos x - \cos \alpha} \]

Factoring out a 2 from the numerator:

\[ \frac{2 (\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} \]

Now, we can factor the numerator further using the difference of squares identity:

\[ \frac{2[(\cos x + \cos \alpha)(\cos x - \cos \alpha)]}{\cos x - \cos \alpha} \]

Canceling out the common factor \((\cos x - \cos \alpha)\) in the numerator and denominator, we have:

\[ 2(\cos x + \cos \alpha) \]

Now we can integrate this expression:

\[ \int 2 (\cos x + \cos \alpha) \, dx \]

Integrating term by term:

\[ 2 \int \cos x \, dx + 2 \int \cos \alpha \, dx \]

The integral of \(\cos x\) with respect to \(x\) is \(\sin x\), and the integral of a constant (\(\cos \alpha\)) with respect to \(x\) is the constant times \(x\):

\[ 2 \sin x + 2 (\cos \alpha) x + c \]

Thus, the integral \(\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx\) is equal to:

\[ 2 (\sin x + x \cos \alpha) + c \]

This corresponds to option (C) \(2 (\sin x + x \cos \alpha) + c\).

Answer 2

We are given: \[ \int \frac{\cos(2x) - \cos(2\alpha)}{\cos(x) - \cos(\alpha)} \, dx \] 

Step 1: Use the identity
\[ \cos(2x) = 2\cos^2(x) - 1,\quad \cos(2\alpha) = 2\cos^2(\alpha) - 1 \] So numerator becomes: \[ 2\cos^2(x) - 1 - (2\cos^2(\alpha) - 1) = 2(\cos^2(x) - \cos^2(\alpha)) \] Use identity: \[ a^2 - b^2 = (a - b)(a + b) \Rightarrow \cos^2(x) - \cos^2(\alpha) = (\cos(x) - \cos(\alpha))(\cos(x) + \cos(\alpha)) \] Thus the integrand becomes: \[ \frac{2(\cos(x) - \cos(\alpha))(\cos(x) + \cos(\alpha))}{\cos(x) - \cos(\alpha)} = 2(\cos(x) + \cos(\alpha)) \] 

Step 2: Integrate
\[ \int 2(\cos(x) + \cos(\alpha)) \, dx = 2 \int \cos(x) \, dx + 2\cos(\alpha) \int dx = 2\sin(x) + 2x\cos(\alpha) + c \] 

Final answer: \(2(\sin(x) + x \cos(\alpha)) + c\)

Answer 3

Let the integral be \(I\).

\[ I = \int \frac{\cos(2x) - \cos(2\alpha)}{\cos(x) - \cos(\alpha)} \, dx \] 

We use the double angle identity for cosine: \(\cos(2\theta) = 2\cos^2(\theta) - 1\).

Applying this identity to the numerator:

\(\cos(2x) - \cos(2\alpha) = (2\cos^2(x) - 1) - (2\cos^2(\alpha) - 1)\)

\(= 2\cos^2(x) - 1 - 2\cos^2(\alpha) + 1\)

\(= 2\cos^2(x) - 2\cos^2(\alpha)\)

\(= 2(\cos^2(x) - \cos^2(\alpha))\)

Now, substitute this back into the integrand:

\[ \frac{\cos(2x) - \cos(2\alpha)}{\cos(x) - \cos(\alpha)} = \frac{2(\cos^2(x) - \cos^2(\alpha))}{\cos(x) - \cos(\alpha)} \]

The term \(\cos^2(x) - \cos^2(\alpha)\) is a difference of squares, which can be factored as \((\cos(x) - \cos(\alpha))(\cos(x) + \cos(\alpha))\).

So the expression becomes:

\[ \frac{2(\cos(x) - \cos(\alpha))(\cos(x) + \cos(\alpha))}{\cos(x) - \cos(\alpha)} \]

Assuming \(\cos(x) \neq \cos(\alpha)\), we can cancel the term \((\cos(x) - \cos(\alpha))\):

\[ = 2(\cos(x) + \cos(\alpha)) \]

Now we can integrate this simplified expression:

\[ I = \int 2(\cos(x) + \cos(\alpha)) \, dx \]

Since \(\alpha\) is a constant, \(\cos(\alpha)\) is also a constant.

\[ I = 2 \left[ \int \cos(x) \, dx + \int \cos(\alpha) \, dx \right] \]

\[ I = 2 \left[ \sin(x) + \cos(\alpha) \int 1 \, dx \right] \]

\[ I = 2 [\sin(x) + x \cos(\alpha)] + c \]

where \(c\) is the constant of integration.

So the integral is \(2(\sin(x) + x \cos(\alpha)) + c\).

Comparing this with the given options, the correct option is:

\(2(\sin(x) + x \cos(\alpha)) + c\)