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Evaluating a definite integral using the arcsine antiderivative

Use the Fundamental Theorem of Calculus, Part 2, along with the fact that arcsin⁡(x)\arcsin(x) is an antiderivative of 11−x2\dfrac{1}{\sqrt{1-x^2}}, to evaluate the definite integral

∫01/211−x2 dx\int_0^{1/2} \frac{1}{\sqrt{1-x^2}}\, dx

What is the exact value of this integral?

Hint

By FTC Part 2, ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) where FF is an antiderivative of ff. Here F(x)=arcsin⁡(x)F(x) = \arcsin(x), so compute arcsin⁡(1/2)−arcsin⁡(0)\arcsin(1/2) - \arcsin(0) using known values of arcsine.

Step-by-step solution

Since arcsin⁡(x)\arcsin(x) is an antiderivative of 11−x2\dfrac{1}{\sqrt{1-x^2}}, FTC Part 2 tells us that ∫01/211−x2 dx=arcsin⁡(x)∣01/2=arcsin⁡(1/2)−arcsin⁡(0)\int_0^{1/2} \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin(x)\Big|_0^{1/2} = \arcsin(1/2) - \arcsin(0).

Recall that arcsin⁡(1/2)=π/6\arcsin(1/2) = \pi/6 (since sin⁡(π/6)=1/2\sin(\pi/6) = 1/2) and arcsin⁡(0)=0\arcsin(0) = 0. Therefore the integral evaluates to π/6−0=π/6\pi/6 - 0 = \pi/6.

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