Let's solve the integral step by step: \[ I = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx \] ### Step 1: Use Substitution Let \( x = t^2 \Rightarrow dx = 2t\,dt \). - When \( x = 0, t = 0 \) - When \( x = 1, t = 1 \) Substitute into the integral: \[ I = \int_{x=0}^{x=1} \frac{\cos x}{\sqrt{x}} dx = \int_{t=0}^{t=1} \frac{\cos(t^2)}{\sqrt{t^2}} \cdot 2t \, dt \] Since \(\sqrt{t^2} = t\) for \(t \geq 0\): \[ I = \int_{0}^{1} \cos(t^2) \cdot \frac{2t}{t} dt = \int_{0}^{1} 2\cos(t^2) dt \] ### Step 2: Use Fresnel Integral The integral \(\int \cos(t^2)\,dt\) does not have an elementary antiderivative, but it is related to the **Fresnel integral** \( C(z) \): \[ C(z) = \int_0^z \cos(\pi t^2/2) dt \] But our limits are from 0 to 1 and the argument is just \(t^2\), not \(\pi t^2/2\). We proceed as follows: Let’s set \(u = t^2\), then \(t = \sqrt{u}\), \(dt = \frac{1}{2\sqrt{u}} du\): \[ \int_0^1 \cos(t^2) dt = \int_{u=0}^{u=1} \cos(u) \cdot \frac{1}{2\sqrt{u}} du \] But this is just going back to our original integral. So let's express the answer in terms of the **Fresnel integral S(x)**: But since we have \(2 \int_0^1 \cos(t^2) dt\), we can just leave the answer as: ### Final Answer \[ \boxed{ I = 2 \int_0^1 \cos(t^2) dt } \] Or, numerically: \[ I \approx 2 \times 0.904524 = 1.80905 \] So, \[ \boxed{I \approx 1.81} \] --- **Summary of Steps:** 1. Substituted \(x = t^2\), \(dx = 2t\,dt\) 2. The integral became \(2 \int_0^1 \cos(t^2) dt\) 3. This is related to the Fresnel cosine integral, and numerically \(I \approx 1.81\).