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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

So

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,

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Tags: Definite Integral, Improper Integral, L'Hôpital's Rule, Math, Monday Math

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