How do you integrate ∫ sin^4 x dx??

2 Answers
May 4, 2018

I=1/32[12x-8sin2x+sin4x]+cI=132[12x8sin2x+sin4x]+c

Explanation:

We know that,

(1)sin^2theta=(1-cos2theta)/2(1)sin2θ=1cos2θ2

(2)cos^2 2theta=(1+cos4theta)/2(2)cos22θ=1+cos4θ2

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

Here,

I=∫ sin^4 x dxI=sin4xdx

=int(sin^2x)^2dx..toApply(1)=(sin2x)2dx..Apply(1)

=int((1-cos2x)/2)^2dx=(1cos2x2)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

May 4, 2018

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

Explanation:

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