sin-2-4x-cos-4x-dx-

Question Number 153760 by ZiYangLee last updated on 10/Sep/21

\int \sin^{2} 4 x \cos 4 x d x =

Answered by puissant last updated on 10/Sep/21

= \int {s i n}^{2} 4 x c o s 4 x d x

= \frac{1}{2} \int s i n 4 x s i n 8 x d x

= - \frac{1}{4} \int - 2 s i n 8 x s i n 4 x d x

= - \frac{1}{4} \int (c o s 12 x + c o s 4 x) d x

= - \frac{1}{4} [\frac{1}{12} s i n 12 x + \frac{1}{4} s i n 4 x] + C

= - \frac{1}{48} s i n 12 x - \frac{1}{16} s i n 4 x + C

∴∵ I = - \frac{1}{48} (s i n 12 x + 3 s i n 4 x) + C . .

Commented by peter frank last updated on 10/Sep/21

thanks

Commented by puissant last updated on 10/Sep/21

{y o u}^{'} r e w e l c o m e