TS EAMCET · Maths · Indefinite Integration
\(\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x\) is equal to
- A \(x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c\)
- B \(2 x+\log (2 \sin x+3 \cos x)+c\)
- C \(x+2 \log (2 \sin x+3 \cos x)+c\)
- D \(\frac{1}{2} \log (4 \sin x+6 \cos x)+c\)
Answer & Solution
Correct Answer
(A) \(x+\frac{1}{2} \log (4 \sin x+6 \cos x)+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x\) We can write \(\sin x+8 \cos x=A(4 \sin x+6 \cos x)\) \(+B \frac{d}{d x}(4 \sin x+6 \cos x)\) \(\sin x+8 \cos x\) \(=A(4 \sin x+6 \cos x)+B(4 \cos x-6 \sin x)\) On equating the coefficient of \(\sin x\) and \(\cos x\),…
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