TS EAMCET · Maths · Indefinite Integration
\(\int \frac{d x}{4+3 \cot x}=\)
- A \(-\frac{3}{25} \log |4+3 \cot x|+\frac{4}{25} x+c\)
- B \(-\frac{3}{25} \log |4 \sin x+3 \cos x|+\frac{4}{25} x+c\)
- C \(\frac{4}{25} \log |4 \sin x+3 \cos x|-\frac{3}{25} x+c\)
- D \(\frac{4}{25} \log |4+3 \cot x|-\frac{3}{25} x+c\)
Answer & Solution
Correct Answer
(B) \(-\frac{3}{25} \log |4 \sin x+3 \cos x|+\frac{4}{25} x+c\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & I=\int \frac{d x}{4+3 \cot x}=\int \frac{\sin x d x}{4 \sin x+3 \cos x} \\ &= \int \frac{A(4 \cos x-3 \sin x)+B(4 \sin x+3 \cos x)}{(4 \sin x+3 \cos x)} d x \\ &= \int\left[\frac{(4 B-3 A) \sin x+(3 B+4 A) \cos x}{4 \sin x+3 \cos x}\right] d x \\ & 3 B+4 A=0…
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