integration of sin⁴xdx
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integration of sin⁴xdx
ʃsin4x dx = ʃsin2x.sin2x dx
= ʃ(1-cos2x)2/4 dx
= ¼ ʃ (cos22x + 1 – 2cos2x )dx
=¼ ʃ [(1+cos4x)/2 + 1 – 2cos2x ]dx
=¼ ʃ [(1+cos4x)/2 dx + ʃ ¼ dx – 2/4 ʃcos2x dx
=⅛ [x + sin4x/4] + x/4 – ½ sin2x/2
=⅛ x + (sin4x)/32 + x/4 – (sin2x)/4
=⅜ x + (sin4x)/32 – (sin2x)/4 + C
ʃsin4x dx = ʃsin2x.sin2x dx
= ʃ(1-cos2x)2/4 dx
= ¼ ʃ (cos22x + 1 – 2cos2x )dx
=¼ ʃ [(1+cos4x)/2 + 1 – 2cos2x ]dx
=¼ ʃ [(1+cos4x)/2 dx + ʃ ¼ dx – 2/4 ʃcos2x dx
=⅛ [x + sin4x/4] + x/4 – ½ sin2x/2
=⅛ x + (sin4x)/32 + x/4 – (sin2x)/4
=⅜ x + (sin4x)/32 – (sin2x)/4 + C
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The integral of the cos is the sin and I get the sin u or sin(4x). So then finally my answer is that the integral of the sin to the fourth of x dx is equal to 3/8 x - 1/4 sin(2x) + 1/32 sin(4x) + a constant C.Mar 15, 2015.
I hope it helps
I hope it helps
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