Question:
$\int \cos 3 x \cos 4 x d x$
Solution:
We know $2 \cos A \cos B=\cos (A-B)+\cos (A+B)$
$\therefore \cos 4 x \cos 3 x=\frac{\cos x+\cos 7 x}{2}$
$\therefore$ The above equation becomes
$\Rightarrow \int \frac{1}{2}(\cos x-\cos 7 x) d x$
$\Rightarrow \frac{1}{2}\left(\int \cos x d x-\int \cos 7 x d x\right)$
We know $\int \cos a x d x=\frac{1}{a} \sin a x+c$
$\Rightarrow \frac{1}{2}\left(\sin x-\frac{1}{7} \sin 7 x\right)$
$\Rightarrow \frac{7 \sin x-\sin 7 x}{14}+C$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.