∫0π2sinx dx\int\limits_{0}^{\frac{\pi}{2}} \sin x \, dx0∫2πsinxdx
= −cosx∣0π2-\cos x \big|_{0}^{\frac{\pi}{2}}−cosx02π
= −cos(π2)−(−cos0)-\cos\left(\frac{\pi}{2}\right) - (-\cos 0)−cos(2π)−(−cos0)
= 0+10 + 10+1
= 1
