Question

\( \int \frac{(5 \sin \theta - 2) \cos \theta}{(5 - \cos^2 \theta - 4 \sin \theta)} d\theta = \)}

• A. \( (\log 5 \sin \theta - 2) + c \)
• B. \( 5 \log(5 \sin \theta - 2) - \frac{8}{(\sin \theta - 2)} + c \)
• C. \( 5 \log |\sin \theta - 2| + \frac{8}{2 - \sin \theta} + c \)
• D. \( \log (5 \sin \theta - 2) + \frac{1}{(\sin \theta - 2)} + c \)

Hint:

When you see $\cos \theta$ in the numerator and only $\sin \theta$ or $\cos^2 \theta$ in the denominator, always substitute $u = \sin \theta$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Use substitution to convert the trigonometric integral into a rational algebraic function.
Step 2: Key Formula or Approach:
Let \( u = \sin \theta \), then \( du = \cos \theta d\theta \).
Step 3: Detailed Explanation:
Convert the denominator first:
\[ 5 - \cos^2 \theta - 4 \sin \theta = 5 - (1 - \sin^2 \theta) - 4 \sin \theta = \sin^2 \theta - 4 \sin \theta + 4 = (\sin \theta - 2)^2 \] Now substitute \( u = \sin \theta \):
\[ \int \frac{5u - 2}{(u - 2)^2} du = \int \frac{5(u - 2) + 8}{(u - 2)^2} du = \int \left( \frac{5}{u-2} + \frac{8}{(u-2)^2} \right) du \] \[ = 5 \log |u - 2| - \frac{8}{u - 2} + c \] Substitute back \( u = \sin \theta \):
\[ = 5 \log |\sin \theta - 2| + \frac{8}{2 - \sin \theta} + c \] Step 4: Final Answer:
The integral is \( 5 \log |\sin \theta - 2| + \frac{8}{2 - \sin \theta} + c \).