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The value of $\int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x}+(\sin x)^{\cos x}} d x$ is
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Verified Answer
The correct answer is:
$\pi / 4$
$I=\int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x}+(\sin x)^{\cos x} d x}$
$I=\int_0^{\pi / 2} \frac{(\sin x)^{\cos x} d x}{(\sin x)^{\cos x}+(\cos x)^{\sin x}}$
$\Rightarrow 2 \mathrm{I}=\pi / 2$
$\Rightarrow I=\pi / 4$
$I=\int_0^{\pi / 2} \frac{(\sin x)^{\cos x} d x}{(\sin x)^{\cos x}+(\cos x)^{\sin x}}$
$\Rightarrow 2 \mathrm{I}=\pi / 2$
$\Rightarrow I=\pi / 4$
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