First, we note that this is an improper integral, as the integrand is not only undefined for the limits of integration, but for the value x=π/4 as well. So, we consider the limit of the integrand as it approaches these values.
First, as x approaches zero (from above) the numerator approaches , while the denominator approaches -∞. Therefore, the integrand goes to zero at the lower limit of integration.
For x=π/4, both and equal zero, and for x=π/2, both approach +∞. Thus, in both cases we have indeterminate forms amenable to L’Hôpital’s rule. And since
and the singularity at x=π/4 is removable, and the (improper) integral is unaffected; and
And the integrand has a finite limit at the upper limit of integration as well. Thus, despite the undefined values, the integrand is sufficiently well-behaved that we need not be much concerned with the improper nature of the integral.
There does not appear to be a clear antiderivative for the integrand, and attempting a substitution like doesn’t appear to help. What we can do, though is an indirect method, which begins with the substitution . Then and . Thus,
But this last integral is just our original I, only with u as the variable of integration instead of x. Therefore,