How do you integrate #int cossqrtx# using integration by parts?

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Answer 1 · 2016-10-27T19:43:28

#intcos(sqrtx)dx=2sqrtxsin(sqrtx)+2cos(sqrtx)+C#

#I=intcos(sqrtx)dx#

We will first use the substitution #t=sqrtx#. This implies that #t^2=x#, which we differentiate to see that #2tdt=dx#. Thus:

#I=intcos(t)(2tdt)=2inttcos(t)dt#

We should now use integration by parts, which takes the form #intudv=uv-intvdu#. For the integral #inttcos(t)dt#, let:

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

So:

#I=2[tsin(t)-intsin(t)dt]#

Since #intsin(t)=-cos(t)+C#:

#I=2[tsin(t)+cos(t)]+C#

Since #t=sqrtx#:

#I=2sqrtxsin(sqrtx)+2cos(sqrtx)+C#