How do you integrate ∫ sin^4 x dx??

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Answer 1 · 2018-05-04T09:34:19

$I=1/32[12x-8sin2x+sin4x]+c$

We know that,

$(1)sin^2theta=(1-cos2theta)/2$

$(2)cos^2 2theta=(1+cos4theta)/2$

$(3)intcosAxdx=1/AsinAx+c$

Here,

$I=∫ sin^4 x dx$

$=int(sin^2x)^2dx..toApply(1)$

$=int((1-cos2x)/2)^2dx$

$=1/4int(1-2cos2x+cos^2 2x)...toApply(2)$

$=1/4int(1-2cos2x+(1+cos4x)/2)dx$

$=1/8int(2-4cos2x+1+cos4x)dx$

$=1/8int(3-4cos2x+cos4x)dx...toApply(3)$

$=1/8[3x-(4sin2x)/2+(sin4x)/4]+c$

$=1/32[12x-8sin2x+sin4x]+c$

Answer 2 · 2018-05-04T09:39:49

The answer is $=1/32sin(4x)-1/4sin(2x)+3/8x+C$

Apply Euler's identity

$sinx=(e^(ix)-e^(-ix))/(2i)$

$cosx=(e^(ix)+e^(-ix))/2$

$i^2=-1$

$sin^4(x)=((e^(ix)-e^(-ix))/(2i))^4$

$=1/16(e^(ix)-e^(-ix))^4$

$=1/16((e^(4ix)+e^(-4ix))-4(e^(2ix)+e^(-2ix))+6)$

$=1/8((e^(4ix)+e^(-4ix))/2-4(e^(2ix)+e^(-2ix))/2+6/2)$

$=1/8(cos4x-4cos2x+3)$

The integral is

$intsin^4xdx=1/8int(cos4x-4cos2x+3)dx$

$=1/32sin(4x)-1/4sin(2x)+3/8x+C$