CUET Preparation Today
$\int\left(\frac{\cos 2x -\cos 2α}{\cos x -cos α}\right)dx=$ |
$2 \sin x + 2x \cos α + c$ $2 \cos x+2x \sin α + c$ $x \cos x + 2 \sin α + c$ $x \cos x + \sin α + c$ |
$2 \sin x + 2x \cos α + c$ |
The correct answer is Option (1) → $2 \sin x + 2x \cos α + c$ \[ \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx = \int \frac{2\cos^2 x - 1 - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} \, dx = \int \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} \, dx \] \[ = \int 2(\cos x + \cos \alpha) \, dx = 2\int \cos x \, dx + 2\cos \alpha \int dx = 2\sin x + 2x\cos \alpha + C \] |
