CUET Preparation Today
$\int\frac{\cos 2x- \cos 2a}{\cos x-\cos a}dx$ is equal to |
$2 \sin x + 2x \cos a +C$, where C is an arbitary constant $2 \sin x + 2 \sin a +C$, where C is an arbitary constant $\frac{\sin 2x}{2}+x \cos 2a + C$, where C is an arbitary constant $\frac{\sin 2x}{2}+\frac{\sin 2a}{2} + C$, where C is an arbitary constant |
$2 \sin x + 2x \cos a +C$, where C is an arbitary constant |
The correct answer is Option (1) → $2 \sin x + 2x \cos a +C$, where C is an arbitary constant $\displaystyle \int \frac{\cos2x-\cos2a}{\cos x-\cos a}\,dx$ $\cos2x-\cos2a=2(\cos^2x-\cos^2a)=2(\cos x-\cos a)(\cos x+\cos a)$ $\displaystyle \int \frac{\cos2x-\cos2a}{\cos x-\cos a}\,dx=\int 2(\cos x+\cos a)\,dx$ $\displaystyle =2\sin x+2x\cos a+C$ |
