AP EAMCET · Maths · Indefinite Integration
\(\int \frac{d x}{12 \cos x+5 \sin x}=\)
- A \(\frac{1}{13} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}-\frac{1}{2} \operatorname{Tan}^{-1} \frac{5}{12}\right)\right|+c\)
- B \(\frac{5}{12} \log \left|\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}-\frac{1}{2} \operatorname{Tan}^{-1} \frac{5}{12}\right)\right|+\mathrm{c}\)
- C \(\frac{1}{13} \log \left|\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}+\frac{1}{2} \operatorname{Tan}^{-1} \frac{5}{12}\right)\right|+\mathrm{c}\)
- D \(\frac{5}{12} \log \left|\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}+\frac{1}{2} \operatorname{Tan}^{-1} \frac{5}{12}\right)\right|+\mathrm{c}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{13} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}-\frac{1}{2} \operatorname{Tan}^{-1} \frac{5}{12}\right)\right|+c\)
Step-by-step Solution
Detailed explanation
\(R = \sqrt{12^2 + 5^2} = \sqrt{144+25} = \sqrt{169} = 13\) \(\alpha = \operatorname{Tan}^{-1} \frac{5}{12}\) \(\int \frac{dx}{12 \cos x + 5 \sin x} = \int \frac{dx}{13 \cos(x - \alpha)} = \frac{1}{13} \int \sec(x - \alpha) dx\)…
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