The integral shown is:

$\int_{0}^{t} \frac{1}{\sqrt{4 - x}} \, dx$

Solution:

To solve this, we use substitution techniques. The denominator suggests the use of a trigonometric substitution.

Step 1: Substitution

Let: $x = 4 \sin^2\theta \quad \text{so that} \quad dx = 8 \sin\theta \cos\theta \, d\theta$ and note that: $4 - x = 4 - 4 \sin^2\theta = 4 \cos^2\theta$ Thus: $\sqrt{4 - x} = 2 \cos\theta$

Step 2: Rewrite the integral

Substituting into the integral: $\int \frac{1}{\sqrt{4 - x}} \, dx = \int \frac{1}{2 \cos\theta} (8 \sin\theta \cos\theta) \, d\theta$ Simplify: $= \int 4 \sin\theta \, d\theta$

Step 3: Integrate

The integral of $4 \sin\theta$ is: $\int 4 \sin\theta \, d\theta = -4 \cos\theta + C$

Step 4: Back-substitute

From the substitution $x = 4 \sin^2\theta$, we know that: $\sin\theta = \sqrt{\frac{x}{4}} = \frac{\sqrt{x}}{2} \quad \text{and} \quad \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{x}{4}} = \sqrt{\frac{4 - x}{4}} = \frac{\sqrt{4 - x}}{2}$ Thus: $\cos\theta = \sqrt{\frac{4 - x}{4}} = \frac{\sqrt{4 - x}}{2}$

Step 5: Evaluate the definite integral

Using the limits $x = 0$ to $x = t$:

• At $x = 0$: $\sin\theta = 0 \implies \theta = 0$
• At $x = t$: $\sin\theta = \frac{\sqrt{t}}{2}$

The integral becomes: $\int_{0}^{t} \frac{1}{\sqrt{4 - x}} \, dx = -4 \cos\theta \bigg|_{0}^{\arcsin\left(\frac{\sqrt{t}}{2}\right)}$

Substitute: $= -4 \left(\frac{\sqrt{4 - t}}{2}\right) + 4 \cdot 1$

Simplify: $\int_{0}^{t} \frac{1}{\sqrt{4 - x}} \, dx = 4 - 2 \sqrt{4 - t}$

Final Answer:

$\int_{0}^{t} \frac{1}{\sqrt{4 - x}} \, dx = 4 - 2\sqrt{4 - t}$

Let me know if you'd like further clarification or step-by-step breakdowns!

Related Questions:

1. How can substitution simplify integrals involving roots?
2. What are the standard trigonometric substitutions for integrals?
3. What is the geometric interpretation of this integral?
4. How would the result change for $\int_{0}^{t} \frac{1}{\sqrt{a - x}} \, dx$?
5. What are some applications of this type of integral in physics or engineering?

Tip:

When solving integrals involving roots or quadratic expressions, always look for opportunities to apply trigonometric substitutions—they often simplify the problem significantly.