Question

(a) Prove that \(\int_{0}^{\pi / 2} \sin ^{2} x d x=\int_{0}^{\pi / 2} \cos ^{2} x d x\) by making the change of variable \(x=\frac{1}{2} \pi-t\) in one of the integrals. (b) Use the same method to prove that the averages of \(\sin ^{2}(n \pi x / l)\) and \(\cos ^{2}(n \pi x / l)\) are the same over a period.

Short answer

\( \int_{0}^{\pi/2} \sin^{2} x \, dx = \int_{0}^{\pi/2} \cos^{2} x \, dx \). The averages of \( \sin^{2} \left( \frac{n \pi x}{l} \right) \) and \( \cos^{2} \left( \frac{n \pi x}{l} \right) \) are equal.

Step-by-step solution

- Define the change of variable

Start by defining the change of variable for the given integral. Let’s use the substitution: \( x = \frac{\pi}{2} - t \).

- Compute the differential

Next, compute the differential of the new variable: \( dx = -dt \).

- Update the limits of integration

Update the limits of integration for the new variable.When \( x = 0 \), \( t = \frac{\pi}{2} \). When \( x = \frac{\pi}{2} \), \( t = 0 \).

- Transform the integral

Transform the given integral \( \int_{0}^{\pi/2} \sin^{2} x \, dx \) with the new variable, limits, and differential:\[ \int_{0}^{\pi/2} \sin^{2} x \, dx = \int_{\pi/2}^{0} \sin^{2} \left( \frac{\pi}{2} - t \right) (-dt) \].

- Simplify the integral

Recognize that \( \sin \left( \frac{\pi}{2} - t \right) = \cos t \). Substitute and simplify:\[ \int_{\pi/2}^{0} \cos^{2} t \, (-dt) = \int_{0}^{\pi/2} \cos^{2} t \, dt \].

- Prove the equality

You have shown that\[ \int_{0}^{\pi/2} \sin^{2} x \, dx = \int_{0}^{\pi/2} \cos^{2} x \, dx \].

- Apply to the given average formula

Using the same method, consider the integrals of \( \sin^{2} \left( \frac{n \pi x}{l} \right) \) and \( \cos^{2} \left( \frac{n \pi x}{l} \right) \) over a period \( \left[ 0, l \right] \). Use the same type of substitution for \( x \):\( x = l - t \).

- Compute differential and transform

Compute \( dx = -dt \) and transform the limits and functions similarly to show that:\[ \int_{0}^{l} \sin^{2} \left( \frac{n \pi x}{l} \right) dx = \int_{0}^{l} \cos^{2} \left( \frac{n \pi x}{l} \right) dx \].

Key concepts

Change of Variables

In calculus, changing variables can simplify an integral, making it easier to solve. A variable change involves substituting a new variable in place of the original one. Here, we changed the variable from \(x\) to \(t\). Our substitution was \(x = \frac{\pi}{2} - t\). Running through this process, we recalculated the limits of integration and adjusted the differential accordingly.

• Original limit: \( x = 0\).
• New limit: \( t = \frac{\pi}{2}\).
• Original limit: \( x = \frac{\pi}{2}\).
• New limit: \( t = 0\).

The differential change involves calculating \( dx = -dt \). Always remember to adjust the limits of integration to match the new variable.

Integration by Substitution

Integration by substitution, a crucial solving technique, helps to transform complex integrals. By substituting \( x = \frac{\pi}{2} - t \), we can change the integral of \( \sin^{2} x \) to a form involving \( \cos \). After substitution, the original integral

• \( \int_{0}^{\pi/2} \sin^{2} x dx\)

becomes

• \( \int_{\pi/2}^{0} \sin^{2} (\frac{\pi}{2} - t) (-dt)\),

and after simplification using the trigonometric identity, turns into

• \( \int_{0}^{\pi/2} \cos^{2} t dt\).

It's vital to include the differential \( (-dt) \) correctly when substituting, which maintains the integral's balance.

Trigonometric Identities

Trigonometric identities are mathematical expressions that relate different trigonometric functions. They are indispensable when simplifying integrals. In the problem above, we use the co-function identity:

• \( \sin (\frac{\pi}{2} - t) = \cos t \).

This identity helped transform \( \sin^{2} (\frac{\pi}{2} - t) \) into \( \cos^{2} t\), making our integral more manageable. Remember, common trigonometric identities include:

• \( \sin^{2} x + \cos^{2} x = 1 \)
• \( \sin (a + b) = \sin a \cos b + \cos a \sin b\)
• \( \cos (a + b) = \cos a \cos b - \sin a \sin b \)

These identities are essential tools in your toolkit when dealing with trigonometric integrals.

Average Value of a Function

The average value of a function provides a measure of the 'mean' level of the function over an interval. In the context of our problem, we aim to show the similarities between the averages of \( \sin^{2}(n \pi x / l) \) and \( \cos^{2}(n \pi x / l) \) over a period. The average value of a function \( f(x) \) over an interval \([a, b]\) is given by:

• \( \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) dx\).

When using our method:

• For \( \sin^{2} \, n \pi x / l \),
• For \( \cos^{2} \, n \pi x / l \),
• After substitution and simplification,

you should find their averages to be identical because the trigonometric identities and the limits of integration harmonize precisely.