TS EAMCET · Maths · Indefinite Integration
\(\int \frac{2 \sin x-3 \cos x}{4 \cos x-3 \sin x} d x=\)
- A \(\frac{1}{25}[17 \log |4 \cos x-3 \sin x|-6 x]+c\)
- B \(\frac{1}{25}[x-18 \log |4 \cos x-3 \sin x|]+c\)
- C \(\frac{1}{25}[\log |4 \cos x-3 \sin x|-18 x]+c\)
- D \(\frac{1}{25}[17 x-6 \log |4 \cos x-3 \sin x|]+c\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{25}[\log |4 \cos x-3 \sin x|-18 x]+c\)
Step-by-step Solution
Detailed explanation
Let \( N = 2 \sin x - 3 \cos x \) and \( D = 4 \cos x - 3 \sin x \). We write \( N = A D + B D' \). Comparing coefficients from \( 2 \sin x - 3 \cos x = A(4 \cos x - 3 \sin x) + B(-4 \sin x - 3 \cos x) \): \( -3A - 4B = 2 \) \( 4A - 3B = -3 \)…
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