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