How do you integrate #(sqrt x) * (cos sqrt x)# ?

Perform the substitution

Let #u=sqrtx#, #=>#, #du=1/(2sqrtx)dx#

The integral is

#I=intsqrtxcos(sqrtx)dx=intu*cosu2udu=2intu^2cosudu#

Perform the integration by parts

#intfg'=fg-intf'g#

#f=u^2#, #=>#, #f'=2u#

#g'=cosu#, #=>#, #g=sinu#

Therefore,

#I=2u^2sinu-4intusinudu#

Perform the integration by parts once more,

#f=u#, #=>#, #f'=1#

#g'=sinu#, #=>#, #g=-cosu#

#intusinudu=-ucosu+intcosudu=-ucosu+sinu#

And finally,

#I=2u^2sinu-4(-ucosu+sinu)=2u^2sinu+4ucosu-4sinu#

#=2xsin(sqrt(x))+4sqrtxcos(sqrtx)-4sin(sqrtx)+C#

#int(sqrt x) * (cos sqrt x)dx# #color(white)(wwwwwwwww# Let #sqrtx = t = x^(1/2)#
#intx/sqrtx* cossqrtx dx##color(white)(wwwwwwwhwwww##=> dt = 1/2*x^(-1/2)dx#
#color(white)(wwwwwwwwwwwwwwwwwwwwww##=> dt = 1/(2sqrtx)dx#
#color(white)(wwwwwwwwwwwwwwwwwwwwww##=>2 dt = 1/(sqrtx)dx#

#=>2 intt^2* cost dt#

#color(white)(wwwwwwwwwwwwwwwwwwwwww#

Here, we'll use Integration by parts,

i.e. #int(uv)dx=uintvdx-int(du/dxintvdx)dx#

#color(white)(wwwwwwwwwwwwwwwwwwwwww#

Integrating #intt^2* cost dt# by parts,

#=>2 [t^2* intcost dt - int(dt^2/dt *intcost dt)dt ]#

#=>2 [t^2* sint - 2int(t *sint)dt ]#

#=>2t^2* sint - 4int(t *sint)dt #

#color(white)(wwwwwwwwwwwwwwwwwwwwww#

Again integrating #int(t *sint)dt # by parts,

#=>2t^2* sint - 4[t*intsintdt - int(dt/dt*intsintdt)dt]#

#=>2t^2* sint - 4[t(-cost)- int(-cost)dt]#

#=>2t^2* sint + 4tcost- 4sint#

Replacing, #t = x^(1/2)#

#=>2x* sinsqrtx + 4sqrtxcossqrtx- 4sinsqrtx +C#

We know that ,(Integration by Parts)

#color(red)(int(u*v)dx=uintvdx-int(u^'intvdx)dx#

Here,

#I=intsqrtx*cossqrtxdx#

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

#:.I=int(t)(cost)(2t)dt#

#=2intt^2costdt#

Applying # "Integration by parts"#

#I=2[t^2(sint)-int(2t)sintdt]+c_1#

#I=2t^2sint-4inttsintdt+c_1#

Again applying integration by parts

#I=2t^2sint-4[t(-cost)-int1(-cost)dt+c_2]+c_1#

#I=2t^2sint-4[t(-cost)+sint+c_3+c_2]+c_1#

#I=2t^2sint+4tcost-4sint-4c_3-4c_2+c_1#

#I=2xsinsqrtx+4sqrtxcossqrtx-4sinsqrtx+C#,

where,#C=c_1-4c_2-4c_3#