TS EAMCET · Maths · Indefinite Integration
If \(\tan \alpha=\frac{4}{3}\), then \(\int \frac{1}{3 \cos x-4 \sin x} d x=\)
- A \(\frac{1}{5} \log \left|\tan \left(\frac{x}{2}+\frac{\alpha}{2}\right)\right|+c\)
- B \(\frac{1}{5} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}+\frac{\alpha}{2}\right)\right|+c\)
- C \(\frac{1}{5} \log \left|\tan \left(\frac{\pi}{4}-\frac{x}{2}-\frac{\alpha}{2}\right)\right|+c\)
- D \(\frac{1}{5} \log |\tan (\sec x+\tan x)|+c\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{5} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}+\frac{\alpha}{2}\right)\right|+c\)
Step-by-step Solution
Detailed explanation
Given \(\tan \alpha=\frac{4}{3}\) and \(\int \frac{1}{3 \cos x-4 \sin x} d x\)…
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