Would like to solve this problem. Did almost half of it but need to use a faster metod like Eulers rule? Can it work ?

int(sinx)^4*(cosx)^2(sinx)4(cosx)2= (sinx)^4(1-sin²x) = sin^4x-sin^6xsin4xsin6x (then what formula can be used to solve it faster)

1 Answer
Jun 15, 2018

1/192(12x+sin6x-3sin4x-3sin2x)+C1192(12x+sin6x3sin4x3sin2x)+C.

Explanation:

To integrate even powers of sin" &/or "cossin &/or cos function/s, we

need to convert the power into multiple angle/s of

sin" &/or "cossin &/or cos function/s.

For this, the following Identities are used :

sin^2x=(1-cos2x)/2, &, cos^2x=(1+cos2x)/2sin2x=1cos2x2,&,cos2x=1+cos2x2.

We have, sin^4xcos^2x=sin^2x*sin^2xcos^2xsin4xcos2x=sin2xsin2xcos2x

=(1-cos2x)/2*1/4(4sin^2xcos^2x)=1/8(1-cos2x)(sin^2 2x)=1cos2x214(4sin2xcos2x)=18(1cos2x)(sin22x),

1/8(1-cos2x)*(1-cos4x)/218(1cos2x)1cos4x2,

=1/16(1-cos4x-cos2x+cos4xcos2x)=116(1cos4xcos2x+cos4xcos2x),

=1/32(2-2cos4x-2cos2x+2cos4xcos2x)=132(22cos4x2cos2x+2cos4xcos2x),

=1/32{2-2cos4x-2cos2x+cos(4x+2x)+cos(4x-2x)}=132{22cos4x2cos2x+cos(4x+2x)+cos(4x2x)},

=1/32(2+cos6x-2cos4x-cos2x)=132(2+cos6x2cos4xcos2x).

:. intsin^4xcos^2xdx,

=1/32{2x+1/6sin6x-2*1/4sin4x-1/2sin2x},

=1/192(12x+sin6x-3sin4x-3sin2x)+C.