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$\int_{0}^{\pi / 2} \frac{d x}{1+\tan x}$ is equal to
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Verified Answer
The correct answer is:
$\pi / 4$
Given, $I=\int_{0}^{\pi / 2} \frac{d x}{1+\tan x}$
$$
\begin{array}{rlr}
I & =\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x & \ldots \\
I & =\int_{0}^{\pi / 2} \frac{\cos (\pi / 2-x)}{\sin (\pi / 2-x)+\cos (\pi / 2-x)} d x \\
& =\int_{0}^{\pi / 2} \frac{\sin x}{\cos x+\sin x} d x & \ldots
\end{array}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 I &=\int_{0}^{\pi / 2}\left(\frac{\sin x+\cos x}{\sin x+\cos x}\right) d x \\
&=\int_{0}^{\pi / 2} d x=\pi / 2 \\
\Rightarrow \quad I &=\pi / 4
\end{aligned}
$$
$$
\begin{array}{rlr}
I & =\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x & \ldots \\
I & =\int_{0}^{\pi / 2} \frac{\cos (\pi / 2-x)}{\sin (\pi / 2-x)+\cos (\pi / 2-x)} d x \\
& =\int_{0}^{\pi / 2} \frac{\sin x}{\cos x+\sin x} d x & \ldots
\end{array}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 I &=\int_{0}^{\pi / 2}\left(\frac{\sin x+\cos x}{\sin x+\cos x}\right) d x \\
&=\int_{0}^{\pi / 2} d x=\pi / 2 \\
\Rightarrow \quad I &=\pi / 4
\end{aligned}
$$
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