Complementary inverses

Calculus Level 4

Evaluate 1729 π 1729.5 π [ sin 1 ( cos x ) + cos 1 ( sin x ) ] d x \displaystyle \int_{1729 \pi}^{1729.5 \pi} \left[ \sin^{-1} (\cos x) +\cos^{-1} ( \sin x ) \right] dx .

π 2 2 \frac{\pi^2}{2} π 2 4 \frac{\pi^2}{4} π 2 8 \frac{\pi^2}{8} π 2 \pi^2

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1 solution

Vincent Moroney
Jun 14, 2018

I = 1729 π 1729.5 π [ arcsin ( cos x ) + arccos ( sin x ) ] d x = 0 π 2 [ arcsin ( cos x ) + arccos ( sin x ) ] d x = 0 π 2 [ arcsin ( sin ( x + π 2 ) ) + arccos ( cos ( x π 2 ) ) ] d x = 0 π 2 2 x d x = π 2 4 \begin{aligned} I = & \int_{1729\pi}^{1729.5\pi} \Big[ \arcsin(\cos x) + \arccos(\sin x) \Big]\, dx\\ = & \int_{0}^{\frac{\pi}{2}} \Big[ \arcsin(\cos x) + \arccos(\sin x) \Big]\, dx\\ = & \int_0^{\frac{\pi}{2}} \Big[ \arcsin(\sin(x+\frac{\pi}{2})) + \arccos(\cos(x-\frac{\pi}{2})) \Big]\, dx\\ = & \int_0^{\frac{\pi}{2}} 2x\, dx\\ = & \boxed{\frac{\pi^2}{4}} \end{aligned}

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