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

1 Answer
mason m
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

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