Evaluate the integral from -pi/4 to 3pi/4 of 1 divided by (3 cos cubed x + sin 2x + sin squared x) dx.

Step-by-step solution

1. 1

   Check the denominator for zeros before attempting any substitution. Write $D(x) = 3\cos^3x+\sin 2x+\sin^2x$. Any tangent half-angle or Weierstrass substitution is worthless if $D$ vanishes inside the interval, because the integral then fails to exist.

2. 2

   Evaluate D at the endpoints and the midpoint. $D\!\left(-\tfrac{\pi}{4}\right) = \tfrac{3\sqrt2}{4}-\tfrac12 \approx 0.5607>0$; $D(0)=3>0$; $D\!\left(\tfrac{\pi}{2}\right) = 0+0+1 = 1>0$; but $D\!\left(\tfrac{3\pi}{4}\right) = -\tfrac{3\sqrt2}{4}-\tfrac12 \approx -1.5607<0$.

3. 3

   Apply the intermediate value theorem. $D$ is continuous and changes sign between $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{4}$, so there is a $c$ in that range with $D(c)=0$. Numerically the single zero is at $c \approx 1.9569768.$ The integrand $\tfrac{1}{D(x)}$ therefore has an infinite discontinuity strictly inside the interval of integration.

4. 4

   Establish that the zero is simple. $D'(c) \approx -3.3131 \neq 0$, so near $c$ the Taylor expansion gives $D(x) \approx D'(c)(x-c)$ and hence $\frac{1}{D(x)} \approx \frac{1}{D'(c)}\cdot\frac{1}{x-c}.$ A simple zero, not a double one, is the worst case for convergence.

5. 5

   Compare with the model integral 1/(x - c). $\displaystyle\int\frac{dx}{x-c} = \ln|x-c|$, which tends to $-\infty$ as $x\to c$. Both one-sided improper integrals diverge, and they diverge to $-\infty$ and $+\infty$, so no principal-value cancellation rescues the ordinary integral.

6. 6

   Conclude. The integral $\int_{-\pi/4}^{3\pi/4}\frac{dx}{3\cos^3x+\sin 2x+\sin^2x}$ diverges; it has no finite value. Only a Cauchy principal value at $x=c$ would be finite, and that must be stated explicitly if it is what the problem wants.