Question

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

Answer 1

`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`