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