Question

How do you evaluate the integral `int sec^2x/(sqrt(1-tanx))dx` from 0 to `pi/4`?

Answer 1

`int_0^(pi//4)sec^2x/sqrt(1-tanx)dx=2`

Explanation:

We have:

`int_0^(pi//4)sec^2x/sqrt(1-tanx)dx`

This is ripe for substitution. If we let `u=1-tanx` then `du=-sec^2xdx`. Before making these substitutions, note that the integral's bounds will change as well.

The bound of `0` becomes `1-tan(0)=1`. The bound of `pi/4` becomes `1-tan(pi/4)=0`.

Thus we see:

`int_0^(pi//4)sec^2x/sqrt(1-tanx)dx=-int_0^(pi//4)(-sec^2x)/sqrt(1-tanx)dx=-int_1^0 1/sqrtudu`

Notice that we can take the negative sign to flip the order of the bounds, just to make things look nicer. Write the square root using fractional and negative exponents.

`-int_1^0 1/sqrtudu=int_0^1u^(-1/2)du=[u^(1/2)/(1/2)]_0^1=[2sqrtu]_0^1=2sqrt1-2sqrt0`

`=2`