What is the answer of #intcos^3(sqrtx)dx#?

2 Answers
Apr 26, 2018

#intcos^3(sqrtx)dx = 3/2(sqrtxsin(sqrtx)+cos(sqrtx))+ 1/18(3usin(3sqrtx)+ cos(3sqrtx))+ C#

Explanation:

Let #u = sqrtx#, the #du = 1/(2sqrtx)dx# or #dx = 2sqrtxdu = 2u du# then:

#intcos^3(sqrtx)dx = 2intucos^3(u)du#

Separate into #cos(u)cos^2(u)#:

#intcos^3(sqrtx)dx = 2intucos(u)cos^2(u)du#

Use the identity #cos^2(u) = 1/2(cos(2u)+1)#:

#intcos^3(sqrtx)dx = intucos(u)(cos(2u)+1)du#

Use the distributive property:

#intcos^3(sqrtx)dx = intucos(u)cos(2u)+ucos(u)du#

Use the identity #cos(A)cos(B) = 1/2(cos(A-B)+cos(A+B)# where #A = 2u and B = u#:

#intcos^3(sqrtx)dx = int 1/2u(cos(u)+ cos(3u))+ucos(u)du#

Combine like terms:

#intcos^3(sqrtx)dx = int 3/2ucos(u)+ 1/2ucos(3u)du#

Separate into two integrals:

#intcos^3(sqrtx)dx = 3/2int ucos(u)du+ 1/2int ucos(3u)du#

Both integrals can be integrated by parts:

#intcos^3(sqrtx)dx = 3/2(usin(u)+cos(u))+ 1/18(3usin(3u)+ cos(3u))+ C#

Reverse the substitution #u = sqrtx#:

#intcos^3(sqrtx)dx = 3/2(sqrtxsin(sqrtx)+cos(sqrtx))+ 1/18(3usin(3sqrtx)+ cos(3sqrtx))+ C#

Apr 26, 2018

#=sqrtx/6(sin3sqrtx+9sinsqrtx)+1/18(cos3sqrtx+27cossqrtx)+C#
Hint:
#cos3x=4cos^3x-3cosx#

#=>4cos^3x=cos3x+3cosx#

#=>cos^3x=1/4(cos3x+3cosx)...to(A)#

Explanation:

We know that,

#(1)intcoskx=1/ksinkx+c#

#(2)intsinkx=1/k(-coskx)+c#

Here, #I=intcos^3(sqrtx)dx#

Let , #sqrtx=t=>x=t^2=>dx=2tdt#

#:.I=intcos^3t*2tdt#

#:.I=2inttcos^3tdt#

Using #(A)# ,we get

#I=2xx1/4int t(cos3t+3cost)dt#

#=1/2inttcos3tdt+3/2inttcostdt#

#"Using "color(blue)"Integration by Parts"#, in both integrals,

#I=1/2[tintcos3tdt-int(d/(dt)(t)intcos3tdt)dt]#

#color(white)(......)+3/2[tcostdt-int(d/(dt)(t)intcostdt)dt]toApply(1)#

#I=1/2[t((sin3t)/3)-int(1)(sin3t)/3dt]#

#color(white)(.....)+3/2[t(sint)-int(1)sintdt]#

#I=1/6[tsin3t-intsin3tdt]+3/2[tsint-intsintdt]#

#=1/6[tsin3t-((-cos3t)/3)]+3/2[tsint-(-cost)]+C#

#=t/6sin3t+1/18cos3t+(3t)/2sint+3/2cost+C,#

#=t/6sin3t+(3t)/2sint+1/18cos3t+3/2cost+C#

#=t/6(sin3t+9sint)+1/18(cos3t+27cost)+C#

Substituting back, #t=sqrtx#

#=sqrtx/6(sin3sqrtx+9sinsqrtx)+1/18(cos3sqrtx+27cossqrtx)+C#