Question

How do you test the improper integral `int sintheta/sqrtcostheta` from `[0,pi/2]` and evaluate if possible?

Answer 1

`2`

Explanation:

`I=int_0^(pi/2)sintheta/sqrtcosthetad theta`

Use the substitution `u=costheta`. This implies that `du=-sinthetad theta`.

When we substitute this into the integral, we will have to transform the bounds by plugging the current ones into `costheta`, so the bound of `0` will become `cos(0)=1` and the bound of `pi/2` will become `cos(pi/2)=0`.

Then:

`I=-int_0^(pi/2)(-sintheta)/sqrtcostheta d theta=-int_1^0 1/sqrtudu`

Flipping the integral's bounds with the negative sign and rewriting the exponent:

`I=int_0^1u^(-1/2)du`

Using the rule `intu^ndu=u^(n+1)/(n+1)+C` but using the FTC to evaluate the integral:

`I=[u^(1/2)/(1/2)]_0^1=[2sqrtu]_0^1=2sqrt1-2sqrt0=2`