Today, we’ll solve a problem in a nice visual way without resorting to too many calculations. The problem involves calculating areas underneath a trig curve. In this post, we will explain why this is true:
$$ \int_0^{\pi} \sin^2(x) \, dx = \frac{\pi}{2} $$
In other words, we would like to calculate the shaded area below (in green). The blue curve is a plot of the graph \( y = \sin^2(x) \) for \( 0 \leq x \leq \pi \).
Consider the graph \( y = \cos^2(x) \) for \( 0 \leq x \leq \pi \) and the shaded area below (in red).
Due to how \( \sin(x) \) and \( \cos(x) \) work, it turns out that the green area is the same as the red area (sine and cosine are complementary). In particular,
$$ \int_0^{\pi} \sin^2(x) \, dx \; = \, \int_0^{\pi} \cos^2(x) \, dx $$
Now, we invoke the Pythagorean Trigonometric Identity:
$$ \cos^2(x) + \sin^2(x) = 1 $$
Note that
$$ \int_0^{\pi} \cos^2(x) \, dx \, + \, \int_0^{\pi} \sin^2(x) \, dx $$
$$= \int_0^{\pi} \cos^2(x) + \sin^2(x) \, dx $$
$$= \int_0^{\pi} dx $$
$$= \pi \; – \; 0 $$
$$= \pi$$
So,
$$ \int_0^{\pi} \cos^2(x) \, dx \, + \, \int_0^{\pi} \sin^2(x) \, dx \; = \; \pi $$
Since \( \int_0^{\pi} \sin^2(x) \, dx \; = \, \int_0^{\pi} \cos^2(x) \, dx \),
$$ 2 \cdot \int_0^{\pi} \sin^2(x) \, dx \; = \, \pi $$
Hence,
$$ \int_0^{\pi} \sin^2(x) \, dx = \frac{\pi}{2} $$
So the green area is \( \pi / 2 \). Remark: The above also shows that the red area is \( \pi / 2 \).
