WBJEE · Maths · Indefinite Integration
If \(\int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x=\alpha\left[\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}\right]+c\), then \(\alpha=\)
- A \(\frac{2}{\mathrm{~b}^{2}}\)
- B \(\frac{2}{a^{2}}\)
- C \(-\frac{2}{b^{2}}\)
- D \(-\frac{2}{\mathrm{a}^{2}}\)
Answer & Solution
Correct Answer
(C) \(-\frac{2}{b^{2}}\)
Step-by-step Solution
Detailed explanation
\(\int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x\) Put \(a+b \cos x=t\) \(-b \sin x d x=d t\)…
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