Question

What is the definite integral of this |sin x| . limit is from -pi/2 to +pi/2?

Answer 1

Answer on Question #51700 - Math - Integral Calculus

Question

What is the definite integral of this $|\sin x|$. limit is from -pi/2 to +pi/2?

Solution

It is known that the antiderivative of $\sin(x)$ is $-\cos(x) + C$, where $C$ is an arbitrary real constant. Next, apply Newton-Leibnitz formula, which gives

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin x \, dx = -\cos x \left|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} - \cos \frac{\pi}{2} - \left(-\cos \left(-\frac{\pi}{2}\right)\right) \right| = -\cos \frac{\pi}{2} + \cos \frac{\pi}{2} = 0,$$

$\cos \left(-\frac{\pi}{2}\right) = \cos \frac{\pi}{2}$, because the cosine function is even.

Answer: 0.

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