COMEDK · Maths · 28. Indefinite Integration
The value of \(\int \frac{1}{1+\cos 8 x} d x\) is
- A \(\frac{\tan 8 x}{8}+C\)
- B \(\frac{\tan 2 x}{8}+C\)
- C \(\frac{\tan 4 x}{8}+C\)
- D \(\frac{\tan 4 x}{4}+C\)
Answer & Solution
Correct Answer
(C) \(\frac{\tan 4 x}{8}+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{1}{1+\cos \& x} d x\) \[ \begin{aligned} &=\int \frac{1}{2 \cos ^{2} 4 x} d x=\frac{1}{2} \int \sec ^{2} 4 x d x \\ &=\frac{1}{2}\left(\frac{\tan 4 x}{4}\right)+C=\frac{1}{8} \tan 4 x+C \end{aligned} \]
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