MCQ
$\int {\left( {\sin x\cos x\cos 2x\cos 4x\cos 8x} \right)dx}$ equal
- A$\frac{{ - 1}}{{128}}\cos 16x + C$
- B$\frac{{1}}{{256}}\cos 16x + C$
- C$\frac{{ - 1}}{{256}}\sin 16x + C$
- ✓$\frac{{ - 1}}{{256}}\cos 16x + C$
$ = \frac{1}{4}\int {\sin } \,4x\cos \,4x\cos \,8x\,dx$
$ = \frac{1}{8}\int {\sin } \,8x\cos \,8x\,dx = \frac{1}{{16}}\int {\sin } \,16x\,dx$
$=\frac{-1}{256} \cos 16 x+C$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.