Question

How do you evaluate the definite integral `int sinsqrtx dx` from `[0, pi^2]`?

Answer 1

`int_0^(pi^2)sin(sqrtx)dx=2pi`

Explanation:

First, let's integrate without the bounds.

`I=intsin(sqrtx)dx`

Make the substitution `t=sqrtx`. This implies that `x=t^2`, which when differentiated shows that `dx=2tdt`. Thus:

`I=intsin(t)(2tdt)=2inttsin(t)dt`

We should then integrate this using integration by parts (IBP), which takes the form `intudv=uv-intvdu`. For the integral we're working with, we should let:

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

Recall that you differentiate `u` and integrate `dv` once you've assigned them their values.

Thus:

`I=2[uv-intvdu]=2[-tcos(t)-int(-cos(t))dt]`

`color(white)I=-2tcos(t)+2intcos(t)dt`

`color(white)I=-2tcos(t)+2sin(t)`

`color(white)I=2sin(sqrtx)-2sqrtxcos(sqrtx)`

Now we can apply the bounds:

`I_B=int_0^(pi^2)sin(sqrtx)dx=[2sin(sqrtx)-2sqrtxcos(sqrtx)]_0^(pi^2)`

`color(white)(I_B)=(2sin(pi)-2picos(pi))-(2sin(0)-2(0)cos(0))`

`color(white)(I_B)=(0-2pi(-1))-(0-0)`

`color(white)(I_B)=2pi`