How do you evaluate this trig integral #int 340cos^4(20x) dx#?

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Answer 1 · 2018-05-15T16:24:32

#I=17/32[240x+8sin(40x)+4sin(80x)]+c#

We know that,

#color(red)((1)cos^2theta=(1+cos2theta)/2#

#color(blue)((2)intcosAxdx=1/AsinAx+c#

Here,

#I=int340cos^4(20x)dx#

Subst. , #20x=u=>x=u/20=>dx=1/20du#

So,

#I=int340cos^4uxx1/20du#

#=17int (cos^2u)^2du...toApply(1)#

#=17int((1+cos2u)/2)^2#

#=17/4int(1+2cos2u+cos^2 2u)du#

#=17/4int[1+2cos2u+(1+cos4u)/2]#

#=17/8int[2+4cos2u+1+cos4u]du#

#=17/8int[3+4cos2u+cos4u]du...toApply(2)#

#=17/8[3u+(4sin2u)/2+(sin4u)/4]+c#

#=17/32[12u+8sin2u+sin4u]+c#

Subst. back , #u=20x#

#I=17/32[12(20x)+8sin2(20x)+4sin4(20x)]+c#

#I=17/32[240x+8sin(40x)+4sin(80x)]+c#