Question

Find the integral:
\[ I = \int \frac{\cos \theta \, d\theta}{\sqrt{3 - 3\sin \theta - \cos^2 \theta}} \]

Answer 1

Step 1: Use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \)
\[ 3 - 3\sin \theta - \cos^2 \theta = 3 - 3\sin \theta - (1 - \sin^2 \theta) = 2 - 3\sin \theta + \sin^2 \theta \] Let \( u = \sin \theta \), so \( du = \cos \theta \, d\theta \)
Step 2: Rewrite the integral: \[ I = \int \frac{du}{\sqrt{u^2 - 3u + 2}} \] Step 3: Complete the square in the denominator:
\[ u^2 - 3u + 2 = \left(u - \frac{3}{2}\right)^2 - \frac{1}{4} \] Step 4: Now the integral becomes: \[ I = \int \frac{du}{\sqrt{\left(u - \frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \] This is of the standard form: \[ \int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C \] Step 5: So, \[ I = \ln\left|u - \frac{3}{2} + \sqrt{(u - \frac{3}{2})^2 - \frac{1}{4}}\right| + C \] Substitute back \( u = \sin \theta \): \[ I = \ln\left|\sin \theta - \frac{3}{2} + \sqrt{2 - 3\sin \theta + \sin^2 \theta}\right| + C \]
Final Answer:
\[ \boxed{I = \ln\left|\sin \theta - \frac{3}{2} + \sqrt{2 - 3\sin \theta + \sin^2 \theta}\right| + C} \]