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 $t=sqrtx$ :

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

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