第10章 10.3

(1)

∫02π2sin⁡x2 dx=2[−2cos⁡x2]02π=−4(−1−1)=8\int_0^{2\pi} 2\sin\frac{x}{2}\,dx = 2\left[-2\cos\frac{x}{2}\right]_0^{2\pi} = -4(-1-1) = 8

(2)

∫π454π(sin⁡x−cos⁡x) dx=[−cos⁡x−sin⁡x]π454π=12+12+12+12=42=22\int_{\frac{\pi}{4}}^{\frac{5}{4}\pi} (\sin x - \cos x)\,dx = \left[-\cos x - \sin x\right]_{\frac{\pi}{4}}^{\frac{5}{4}\pi} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}

(3)

∫π6π3tan⁡x dx=[−log⁡∣cos⁡x∣]π6π3=−log⁡12+log⁡32=log⁡3\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \tan x\,dx = \left[-\log|\cos x|\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = -\log\frac{1}{2} + \log\frac{\sqrt{3}}{2} = \log\sqrt{3}

(4)

∫0π4(tan⁡x−sin⁡x) dx=[−log⁡∣cos⁡x∣+cos⁡x]0π4=−log⁡12+12−1=log⁡2+2−22\int_0^{\frac{\pi}{4}} (\tan x - \sin x)\,dx = \left[-\log|\cos x| + \cos x\right]_0^{\frac{\pi}{4}} = -\log\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1 = \log\sqrt{2} + \frac{\sqrt{2}-2}{2}

解説: r31bn1z