AP EAMCET · Maths · Indefinite Integration
\(\int \frac{1}{9 \cos ^2 x-24 \sin x \cos x+16 \sin ^2 x} d x=\)
- A \(\frac{\cos x}{4(3 \cos x-4 \sin x)}+c\)
- B \(\frac{\sin x}{4(3 \cos x-4 \sin x)}+c\)
- C \(\frac{\cos x}{3 \cos x-4 \sin x}+c\)
- D \(\frac{\sin x}{3 \cos x-4 \sin x}+c\)
Answer & Solution
Correct Answer
(A) \(\frac{\cos x}{4(3 \cos x-4 \sin x)}+c\)
Step-by-step Solution
Detailed explanation
\( \int \frac{1}{(3 \cos x - 4 \sin x)^2} d x \) \( = \int \frac{\sec^2 x}{(3 - 4 \tan x)^2} d x \) Let \( u = 3 - 4 \tan x \Rightarrow d u = -4 \sec^2 x d x \) \( = -\frac{1}{4} \int u^{-2} d u \) \( = -\frac{1}{4} \left( -u^{-1} \right) + C \) \( = \frac{1}{4u} + C \)…
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