Question

Use the double-angle formulas to evaluate the following integrals. \(\int_{0}^{\pi} \sin ^{2} x d x\)

Short answer

The integral \(\int_{0}^{\pi} \sin^2 x \, dx\) evaluates to \(\frac{\pi}{2}\).

Step-by-step solution

Identify the Double-Angle Formula for Sine Function

To evaluate the integral \( \int_{0}^{\pi} \sin^2 x \, dx \), we first need to use the double-angle identity. The double-angle formula for \( \sin^2 x \) is \( \sin^2 x = \frac{1 - \, \cos(2x)}{2} \). This identity will allow us to rewrite the integral in a simpler form.

Substitute the Double-Angle Formula into the Integral

Substitute the identity for \( \sin^2 x \) into the integral. The integral becomes: \[ \int_{0}^{\pi} \sin^2 x \, dx = \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx. \]This simplifies our problem into a basic integral where we will integrate each term separately.

Split the Integral into Two Parts

Split the integral into two separate integrals for easier computation:\[ \int_{0}^{\pi} \frac{1}{2} \, dx - \int_{0}^{\pi} \frac{\cos(2x)}{2} \, dx. \]Each of these integrals can be evaluated individually.

Evaluate the First Integral

Evaluate the first integral:\[ \int_{0}^{\pi} \frac{1}{2} \, dx = \frac{1}{2} \left[ x \right]_{0}^{\pi} = \frac{1}{2} (\pi - 0) = \frac{\pi}{2}. \]

Evaluate the Second Integral

Evaluate the second integral using the substitution \( u = 2x \), \( du = 2 \, dx \), hence \( dx = \frac{1}{2} du \):\[ \int_{0}^{\pi} \frac{\cos(2x)}{2} \, dx = \frac{1}{2} \int_{0}^{2\pi} \cos(u) \, \frac{1}{2} \, du = \frac{1}{4} \int_{0}^{2\pi} \cos(u) \, du. \]The integral of \( \cos(u) \) is \( \sin(u) \), so we have:\[ \frac{1}{4} \left[ \sin(u) \right]_{0}^{2\pi} = \frac{1}{4} (\sin(2\pi) - \sin(0)) = \frac{1}{4} (0 - 0) = 0. \]

Combine the Results

Adding the results of the two integrals, we get:\[ \int_{0}^{\pi} \sin^2 x \, dx = \frac{\pi}{2} + 0 = \frac{\pi}{2}. \]

Key concepts

Double-Angle Formulas

Double-angle formulas are crucial in calculus, especially for simplifying expressions involving trigonometric functions. These identities allow us to express trigonometric functions of double angles in terms of single angles. In the given problem, we see the application of the double-angle formula for sine, which is:

• \( \sin^2 x = \frac{1 - \cos(2x)}{2} \)

This identity transforms the integral into a form that is more manageable to solve. Instead of directly integrating \( \sin^2 x \), we use this transformation to reduce it to a simpler expression involving cosine, making the integration process much easier.
Understanding and applying double-angle formulas is key to solving integrals that initially seem complex. They help break down expressions into recognizable components we can easily work with, avoiding complicated and lengthy algebraic manipulations.
Remember, these formulas not only simplify integration but also aid in solving trigonometric equations and proving other trigonometric identities.

Trigonometric Identities

Trigonometric identities form the backbone of solving calculus problems involved with angles and periodic functions. These identities include equations that hold true for any angle and are integral in simplifying expressions before integration or differentiation.
In our context:

• We used a trigonometric identity to convert \( \sin^2 x \) into a function of cosine: \( \sin^2 x = \frac{1 - \cos(2x)}{2} \).

Such identities are essential tools in calculus.

• They simplify complex trigonometric expressions.
• They reduce our calculations by turning tools which may not directly integrate into those that easily integrate.

Memorizing a core set of trigonometric identities enables quick transformations of trigonometric functions, reducing complexity and leading to accurate solutions more efficiently.
By mastering these identities, we can confidently tackle a broad range of calculus problems, from simple integral evaluations to complex limit calculations.

Integral Calculus

Integral calculus is the branch of calculus focused on accumulation of quantities and the area under and between curves. It is widely applied to solve problems involving continuous change and can be used in various fields, from engineering to economics.
The problem at hand involved solving the integral \( \int_{0}^{\pi} \sin^2 x \, dx \) using a double-angle identity. This teaches us a key strategy in integral calculus when faced with awkward integrands: transform them using identities or substitutions into something easily integrable.

• By substituting \( \sin^2 x \) with \( \frac{1 - \cos(2x)}{2} \), the problem simplifies.
• The integration splits into two parts: a basic \( x \) integral and a \( \cos(2x) \) integral.
• We evaluate these individually to obtain our result \( \frac{\pi}{2} \).

Integral calculus is not just about skillfully handling the calculus operations but about understanding and applying strategies, like using identities and substitutions, to simplify complex integrals into manageable forms. Mastering this approach enables students to tackle tougher calculus problems with confidence and ease.