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#