Q106 / 180 · COMEDK 2017 · Mathematics

$\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}$ is equal to

  • Correct answer
    A
    1
  • B
    2
  • C
    3
  • D
    4

Solution

$f(x)=1+\cos 2 x$ and $\begin{aligned} f(-x) &=1+\cos (-2 x)=1+\cos 2 x=f(x) \\ \Rightarrow \quad f(x) &=f(-x) \end{aligned}$ Hence, $f(x)$ is even function. $\begin{aligned} \Rightarrow &=2 \int_{0}^{\pi / 4} \frac{d x}{1+\cos 2 x} \\ &=2 \int_{0}^{\pi / 4} \frac{d x}{2 \cos ^{2} x} \\ &=2 \int_{0}^{\pi / 4} \frac{1}{2} \sec ^{2} x d x \\ &=[\tan x]_{0}^{\pi / 4}=\left(\tan \frac{\pi}{4}-\tan 0\right)=1 \end{aligned}$

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

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