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Answer 1 · 2017-10-28T15:32:17

The answer is #=1/32sin4(x-3)-1/4sin2(x-3)+3/8x+C#

We need

#intcosnxdx=1/nsinnx+C#

Before starting, we calculate #sin^4a# in terms of #cos2a# and #cos4a#

#cos2a=1-2sin^2a#, #=>#, #sin^2 a=(1-cos2a)/2#

#sin^4a=((1-cos2a)/2)^2=1/4(1-2cos2a+cos^2 2a)#

#cos4a=2cos^2 2a-1#, #=>#, #cos^2 2a=(1+cos4a)/2#

Therefore,

#sin^4a=1/4(1-2cos2a+(1+cos4a)/2)#

#=1/8(cos4a-4cos2a+3)#

So,

#int(sin(x-3))^2dx=1/8int(cos4(x-3)-4cos2(x-3)+3)dx#

#=1/8(1/4sin4(x-3)-4/2sin2(x-3)+3x)+C#

#=1/32sin4(x-3)-1/4sin2(x-3)+3/8x+C#

Answer 2 · 2017-10-28T15:35:57

#int sin^4(x-3) dx = 1/4sin^3(x-3)cos(x-3)+ 3/8(x-3 - sin(x-3)cos(x-3))+ C#

Given: #int sin^4(x-3) dx#

Let's make the argument easy to handle:

Let #u = x - 3#, then #du = dx#:

#int sin^4(u) du#

#int sin^n(u) du = 1/nsin^(n-1)(u)cos(u)+ (n-1)/nintsin^(n-2)(u)du#

where #n = 4#

#int sin^4(u) du = 1/4sin^3(u)cos(u)+ 3/4intsin^2(u)du#

The last integral is in the above list of integrals:

#int sin^4(u) du = 1/4sin^3(u)cos(u)+ 3/8(u - sin(u)cos(u))+ C#

Reverse the substitution:

#int sin^4(x-3) dx = 1/4sin^3(x-3)cos(x-3)+ 3/8(x-3 - sin(x-3)cos(x-3))+ C#