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$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1 + \tan^3 x} = ?$
A$\pi/4$
B$\pi/2$
C$0$
D$\ln 2$
Answer & Solution
Correct answer: A. $\pi/4$
Same King's trick as $\sin/(\sin+\cos)$: $1/(1 + \tan^3 x) = \cos^3 x/(\cos^3 x + \sin^3 x)$. Sub $x \to \pi/2 - x$ swaps $\sin↔\cos$: $I = \int \sin^3 x/(\sin^3 x + \cos^3 x)$. Add: $2I = \int_0^{\pi/2} dx = \pi/2$. So $I = \pi/4$. (Indeed, $\int_0^{\pi/2} dx/(1 + \tan^n x) = \pi/4$ for ANY $n$ — striking invariance.)
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