cos-6-x-dx-

Question Number 7898 by tawakalitu last updated on 23/Sep/16

\int {c o s}^{6} x d x

Commented by sandy_suhendra last updated on 24/Sep/16

{c o s}^{6} x = {({c o s}^{2} x)}^{3}

= {(\frac{1}{2} + \frac{1}{2} c o s 2 x)}^{3}

= \frac{1}{8} + \frac{3}{8} c o s 2 x + \frac{3}{8} {c o s}^{2} 2 x + \frac{1}{8} {c o s}^{3} 2 x

= \frac{1}{8} + \frac{3}{8} c o s 2 x + \frac{3}{8} (\frac{1}{2} + \frac{1}{2} c o s 4 x) + \frac{1}{8} c o s 2 x ({c o s}^{2} 2 x)

= \frac{1}{8} + \frac{3}{8} c o s 2 x + \frac{3}{16} + \frac{3}{16} c o s 4 x + \frac{1}{8} c o s 2 x (1 - {s i n}^{2} 2 x)

= \frac{5}{16} + \frac{3}{8} c o s 2 x + \frac{3}{16} c o s 4 x + \frac{1}{8} c o s 2 x - \frac{1}{8} c o s 2 x . {s i n}^{2} 2 x

= \frac{5}{16} + \frac{1}{2} c o s 2 x + \frac{3}{16} c o s 4 x - \frac{1}{8} c o s 2 x . {s i n}^{2} 2 x

\int (\frac{5}{16} + \frac{1}{2} c o s 2 x + \frac{3}{16} c o s 4 x - \frac{1}{8} c o s 2 x . {s i n}^{2} 2 x) d x

= \frac{5}{16} x + \frac{1}{4} s i n 2 x + \frac{3}{64} s i n 4 x - \frac{1}{48} {s i n}^{3} 2 x + c

Commented by tawakalitu last updated on 24/Sep/16

T h a n k s f o r y o u r h e l p .

Answered by prakash jain last updated on 02/Oct/16