Integral Calculation Exercise 4

We aim to evaluate the following definite integral:

$$ \int_0^{\frac{1}{2}} \sqrt{1 - x^2} \ dx $$

The integrand $ \sqrt{1 - x^2} $ resembles the identity $ \cos(x) = \sqrt{1 - \sin^2(x)} $, which follows from the Pythagorean identity.

Note. From the fundamental identity $$ \cos^2 \alpha + \sin^2 \alpha = 1, $$ we can derive: $$ \cos^2 \alpha = 1 - \sin^2 \alpha $$ and thus $$ \sqrt{ \cos^2 \alpha } = \sqrt{ 1 - \sin^2 \alpha } $$ leading to $$ \cos \alpha = \sqrt{ 1 - \sin^2 \alpha } $$

We'll now apply the substitution method by setting $ x = \sin(t) $:

$$ x = \sin(t) $$

Since the sine function is invertible on the interval considered, we can express $ t $ as an arcsine:

$$ t = \arcsin(x) $$

We proceed by differentiating both sides:

$$ D[x] = D[\sin(t)] \quad \Rightarrow \quad dx = \cos(t) \ dt $$

We now substitute dx = cos(t) dt into the original integral:

$$ \int_0^{\frac{1}{2}} \sqrt{1 - x^2} \ dx = \int_0^{\frac{1}{2}} \sqrt{1 - x^2} \cdot \cos(t) \ dt $$

Substituting $ x = \sin(t) $, we get:

$$ \int_0^{\frac{1}{2}} \sqrt{1 - \sin^2(t)} \cdot \cos(t) \ dt $$

Using the identity $ \sqrt{1 - \sin^2(t)} = \cos(t) $, the integrand simplifies to:

$$ \int_0^{\frac{1}{2}} \cos^2(t) \ dt $$

Since the variable of integration is now $ t = \arcsin(x) $, we need to adjust the limits of integration accordingly:

$$ \int_{\arcsin(0)}^{\arcsin(\frac{1}{2})} \cos^2(t) \ dt = \int_0^{\frac{\pi}{6}} \cos^2(t) \ dt $$

We now apply the double-angle identity for cosine: $ \cos(2t) = 2\cos^2(t) - 1 $, which rearranges to $ \cos^2(t) = \frac{\cos(2t) + 1}{2} $.

Using this, we rewrite the integral as:

$$ \int_0^{\frac{\pi}{6}} \frac{\cos(2t) + 1}{2} \ dt $$

Splitting the integral:

$$ \int_0^{\frac{\pi}{6}} \frac{1}{2} \ dt + \int_0^{\frac{\pi}{6}} \frac{\cos(2t)}{2} \ dt $$

$$ \frac{1}{2} \int_0^{\frac{\pi}{6}} dt + \frac{1}{2} \int_0^{\frac{\pi}{6}} \cos(2t) \ dt $$

The first integral is immediate:

$$ \frac{1}{2} [t]_0^{\frac{\pi}{6}} + \frac{1}{2} \int_0^{\frac{\pi}{6}} \cos(2t) \ dt = \frac{\pi}{12} + \frac{1}{2} \int_0^{\frac{\pi}{6}} \cos(2t) \ dt $$

To handle the second integral, we multiply and divide by 2:

$$ \frac{\pi}{12} + \frac{1}{2} \cdot \frac{2}{2} \int_0^{\frac{\pi}{6}} \cos(2t) \ dt = \frac{\pi}{12} + \frac{1}{4} \int_0^{\frac{\pi}{6}} 2 \cos(2t) \ dt $$

Since the antiderivative of $ 2\cos(2t) $ is $ \sin(2t) $, we obtain:

$$ \frac{\pi}{12} + \frac{1}{4} \left[ \sin(2t) \right]_0^{\frac{\pi}{6}} $$

$$ \frac{\pi}{12} + \frac{1}{4} \left( \sin\left(\frac{\pi}{3}\right) - \sin(0) \right ) $$

$$ \frac{\pi}{12} + \frac{1}{4} \cdot \sin\left(\frac{\pi}{3}\right) $$

Since the sine of $ \pi/3 $ is $ \frac{\sqrt{3}}{2} $, this becomes:

$$ \frac{\pi}{12} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{12} + \frac{\sqrt{3}}{8} $$

Therefore, the value of the definite integral is:

$$ \frac{2\pi + 3\sqrt{3}}{24} $$

And that completes the solution.