To integrate the given trigonometric function we have to think about the double identity since:
color(green)(sin^4(4x))=(color(red)(sin^2(4x)))^2
The double trigonometric identity says:
color(blue)((cos2alpha)=1-2sin^2(alpha))
color(blue)(rArr(cos2alpha-1)/-2=sin^2(alpha))
color(blue)(rArr(1-cos2alpha)/2=sin^2(alpha))
Applying this identity on the above equation we have:
color(red)(sin^2(4x)=(1-cos2(4x))/2=(1-cos8x)/2)
color(green)(sin^4(4x))=(color(red)(sin^2(4x)))^2
color(green)(sin^4(4x))=color(red)(((1-cos8x)/2)^2)
color(green)(sin^4(4x))=1/4color(brown)((1-cos8x)^2)
Applying the polynomial identity(square of the difference) that says:
(a-b)^2=a^2-2ab+b^2
color(brown)((1-cos8x)^2=1-2cos8x+cos^2(8x))
color(green)(sin^4(4x))=1/4(color(brown)(1-2cos8x+color(blue)(cos^2(8x))))
Since we still have a power trigonometry we should get rid of this color(brown)(cos^2(8x)) by also using the double trigonometric identity that says:
color(blue)((cos2alpha)=2cos^2(alpha)-1)
color(blue)(rArr(cos2alpha+1)/2=cos^2(alpha))
Here we have,
color(blue)((cos2(8x)+1)/2=cos^2(8x))
color(blue)((cos16x+1)/2=cos^2(8x))
color(green)(sin^4(4x))=1/4(color(brown)(1-2cos8x+color(blue)(cos^2(8x))))
color(green)(sin^4(4x))=1/4(1-2cos8x+color(blue)((cos16x+1)/2))
color(green)(sin^4(4x))=1/4(1-2cos8x+(cos16x)/2+1/2)
color(green)(sin^4(4x))=1/4(3/2-2cos8x+(cos16x)/2)
color(green)(sin^4(4x))=3/8-(cos8x)/2+(cos16x)/8
Now Let us integrate it:
color(green)(int(sin^4(4x))dx)=int(3/8-(cos8x)/2+(cos16x)/8)dx
color(green)(int(sin^4(4x))dx)=int3/8dx-int(cos8x)/2do int(cos16x)/8dx
color(green)(int(sin^4(4x))dx)=(3/8)x-(sin8x)/16+(sin16x)/128+C
