I tried to solve for the area of a quarter circle of radius $4$ using the following integral:
$$\int_{0}^{4} \sqrt{16-x^2} dx$$
Of course, the answer is $4\pi$, but I wanted to get there myself, so after a bit of work I get
$$8\int_0^{\pi/2}(\cos2x+1)\ dx$$
I couldn't proceed from here, so I Googled the question and arrived at this Stack Exchange page. But the second answer just says (from an anonymous user), "As $\cos2u$ runs from $1$ to $-1$ symmetrically, this contribution vanishes and $4\pi$ remain". I don't know what this means though, or where he got the $4\pi$. Could someone please explain?
The guy is the comment (Ted) already answered it for ya... But here are the steps:
Given the integral: $$\int_{0}^{4} \sqrt{16-x^2} \, dx$$
This integral represents the area under the curve (y = $\sqrt{16-x^2}\ $) from (x = 0) to (x = 4). The curve is the upper half of a circle with a radius of 4.
Step 1: Change of Variables (Trigonometric Substitution)
Let's make a trigonometric substitution to simplify the integrand.
Let: [x = 4 $ \sin(\theta)\ $] Then: [dx = 4 $ \cos(\theta) \ $, $ d\theta\ $]
When $x = 0$, ($ \sin(\theta) = 0\ $) which implies $\theta = 0$. When (x = 4), ($ \sin(\theta) = 1\ $) which implies ($ \theta = \frac{\pi}{2}\ $).
Step 2: Substituting the Values
Substitute the values of $x$ and $dx$ into the integral: $$\int_{0}^{4} \sqrt{16-x^2} \, dx = \int_{0}^{\pi/2} \sqrt{16-16\sin^2(\theta)} \cdot 4\cos(\theta) \, d\theta$$
Using the trigonometric identity: $\cos^2(\theta) = 1 - \sin^2(\theta)$
We get: $$\int_{0}^{\pi/2} \sqrt{16\cos^2(\theta)} \cdot 4\cos(\theta) \, d\theta = \int_{0}^{\pi/2} 4\cos(\theta) \cdot 4\cos(\theta) \, d\theta$$
Step 3: Solve the Integral
$$\int_{0}^{\pi/2} 16\cos^2(\theta) \, d\theta$$
Now, we'll use another trigonometric identity to simplify ($ \cos^2(\theta)\ $):
($ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\ $)
Substituting in: $$\int_{0}^{\pi/2} 16 \cdot \frac{1 + \cos(2\theta)}{2} \, d\theta = 8 \int_{0}^{\pi/2} (1 + \cos(2\theta)) \, d\theta$$
Splitting the integral: $$8 \int_{0}^{\pi/2} d\theta + 8 \int_{0}^{\pi/2} \cos(2\theta) \, d\theta$$
The second integral is zero since the integral of ($ \cos(2\theta)\ $) over a complete cycle (from 0 to ($ \pi/2\ $)) is zero.
Therefore, the integral becomes: $$8 \int_{0}^{\pi/2} d\theta = 8[\theta]_{0}^{\pi/2} = 8 \cdot \frac{\pi}{2} = 4\pi$$
So: $$\int_{0}^{4} \sqrt{16-x^2} \, dx = 4\pi$$
This result makes sense geometrically, as the integral represents a quarter of the area of a circle with radius 4, which is ($ \frac{1}{4} \times \pi \times 4^2 = 4\pi \ $).