How do you evaluate the definite integral #int 4/sin^2(z) dz# from #[pi/6, pi/2]#?

Question

Charles Arocha · Accepted Answer

#4sqrt3#.

#int_(pi/6)^(pi/2) 4/sin^2 zdz=4int_(pi/6)^(pi/2)csc^2 zdz# #=-4[cot z]_(pi/6)^(pi/2)# #=-4[cot (pi/2)-cot (pi/6)]# #=-4[0-sqrt3]# #=4sqrt3#.

Adam Adkins · Answer

undefined To evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{4}{\sin^2(z)} \, dz$, we can use the properties of integrals and trigonometric identities to simplify the expression and then evaluate it.

First, we rewrite the integrand using the identity $\sin^2(z) = 1 - \cos^2(z)$.

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{4}{\sin^2(z)} \, dz = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{4}{1 - \cos^2(z)} \, dz$$

Now, we can use the substitution $u = \cos(z)$, so $du = -\sin(z) \, dz$.

$$\int \frac{-4}{1 - u^2} \, du$$

This is a standard integral that can be evaluated using partial fraction decomposition. After performing the decomposition and integrating, we obtain:

$$\int \frac{-4}{1 - u^2} \, du = -2 \ln \left| \frac{1 + u}{1 - u} \right| + C$$

Now, we reverse the substitution:

$$-2 \ln \left| \frac{1 + \cos(z)}{1 - \cos(z)} \right| + C$$

Finally, we evaluate the integral from $\frac{\pi}{6}$ to $\frac{\pi}{2}$:

$$-2 \ln \left| \frac{1 + \cos(\frac{\pi}{2})}{1 - \cos(\frac{\pi}{2})} \right| - (-2 \ln \left| \frac{1 + \cos(\frac{\pi}{6})}{1 - \cos(\frac{\pi}{6})} \right|)$$

Simplify the trigonometric terms:

$$-2 \ln \left| \frac{1}{0} \right| - (-2 \ln \left| \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}} \right|)$$

The natural logarithm of 0 is undefined, so the integral is also undefined.