Q115 / 180 · COMEDK 2020 · Mathematics

The value of $\int \frac{1}{1+\cos 8 x} d x$ is

  • A
    $\frac{\tan 8 x}{8}+C$
  • B
    $\frac{\tan 2 x}{8}+C$
  • Correct answer
    C
    $\frac{\tan 4 x}{8}+C$
  • D
    $\frac{\tan 4 x}{4}+C$

Solution

Let $I=\int \frac{1}{1+\cos \& x} d x$ $$ \begin{aligned} &=\int \frac{1}{2 \cos ^{2} 4 x} d x=\frac{1}{2} \int \sec ^{2} 4 x d x \\ &=\frac{1}{2}\left(\frac{\tan 4 x}{4}\right)+C=\frac{1}{8} \tan 4 x+C \end{aligned} $$

Hence, the correct answer is option: 'C'.

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