Question

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

Answer 1

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

Explanation:

`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`