How do you integrate #int x cos sqrtx dx # using integration by parts?

1 Answer
Nov 27, 2016

#2(sqrtxsin(sqrtx)(x-6)+3cos(sqrtx)(x-2))+C#

Explanation:

Before using integration by parts, let #t=sqrtx#. This implies that #t^2=x#. Differentiating this shows that #2tdt=dx#.

So:

#I=intxcos(sqrtx)dx=intt^2cos(t)(2tdt)=int2t^3cos(t)dt#

Now we should apply integration by parts. IBP takes the form #intudv=uv-intvdu#. So, for #int2t^3cos(t)dt#, let:

#{(u=2t^3,=>,du=6t^2dt),(dv=cos(t)dt,=>,v=sin(t)):}#

Then:

#I=uv-intvdu=2t^3sin(t)-int6t^2sin(t)dt#

For #int6t^2sin(t)dt#, use IBP again:

#{(u=6t^2,=>,du=12tdt),(dv=sin(t)dt,=>,v=-cos(t)):}#

Now:

#I=2t^3sin(t)-[-6t^2cos(t)+int12tcos(t)dt]#

#I=2t^3sin(t)+6t^2cos(t)-int12tcos(t)dt#

Reapplying IBP:

#{(u=12t,=>,du=12dt),(dv=cos(t)dt,=>,v=sin(t)):}#

#I=2t^3sin(t)+6t^2cos(t)-[12tsin(t)-int12sin(t)dt]#

#I=2t^3sin(t)+6t^2cos(t)-12tsin(t)+int12sin(t)dt#

Since #intsin(t)dt=-cos(t)#:

#I=2t^3sin(t)+6t^2cos(t)-12tsin(t)-12cos(t)#

Factoring:

#I=2(tsin(t)(t^2-6)+3cos(t)(t^2-2))#

Since #t=sqrtx#:

#I=2(sqrtxsin(sqrtx)(x-6)+3cos(sqrtx)(x-2))+C#