Question

\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to 

Options

• \[2\left( \sin x + x\cos\theta \right) + C\]

• \[2\left( \sin x - x\cos\theta \right) + C\]

• \[2\left( \sin x + 2x\cos\theta \right) + C\]

• \[2\left( \sin x - 2x\cos\theta \right) + C\]

Answer 1

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

 

\[\text{Let }I = \int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\]

\[ = \int\frac{\left( 2 \cos^2 x - 1 \right) - \left( 2 \cos^2 \theta - 1 \right)}{\cos x - \cos\theta}dx\]

\[ = \int\frac{2 \cos^2 x - 1 - 2 \cos^2 \theta + 1}{\cos x - \cos\theta}dx\]

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

\[ = \int2\left( \cos x + \cos\theta \right)dx\]

\[ = 2\left( \sin x + x\cos\theta \right) + C\]

\[\text{Therefore, }\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx = 2\left( \sin x + x\cos\theta \right) + C\]