sin-8-x-cos-8-x-dx-

Question Number 97041 by bemath last updated on 06/Jun/20

\int \sin^{8} (x) \cos^{8} (x) d x = ?

Answered by john santu last updated on 06/Jun/20

\Rightarrow \sin^{8} x . \cos^{8} x = \frac{\sin^{8} (2 x)}{2^{8}}

\sin (2 x) = \frac{e^{2 i x} - e^{- 2 i x}}{2 i}

\sin^{8} x . \cos^{8} x = \frac{{(e^{2 i x} - e^{- 2 i x})}^{8}}{2^{16}}

I = \frac{1}{2^{15}} \int (\cos 16 x + 8 \cos 12 x + 56 \cos 4 x + 35) d x

I = \frac{1}{2^{15}} (\frac{\sin 16 x}{16} + \frac{8 \sin 12 x}{12} + \frac{56 \sin 4 x}{4} + 35 x) + c

Answered by Sourav mridha last updated on 06/Jun/20

l e t s i n x = m

\int {(1 - m^{2})}^{7} m^{8} d m

= \int [\underset{r = 0}{\sum^{7}} {\overset{7}{C}}_{r} {(1)}^{7 - r} . {(- m^{2})}^{r} .] . m^{8} d m

= \underset{r = 0}{\sum^{7}} {(- 1)}^{r} {\overset{7}{C}}_{r} [\int m^{2 r + 8} d m]

= \underset{r = 0}{\sum^{7}} {(- 1)}^{r} {\overset{7}{C}}_{r} \frac{{(s i n (x))}^{2 r + 9}}{2 r + 9} . + k