Question

Consider the integral \(I=\int \sin ^{2} x \cos ^{2} x d x\) a. Find \(I\) using the identity \(\sin 2 x=2 \sin x \cos x\) b. Find \(I\) using the identity \(\cos ^{2} x=1-\sin ^{2} x\) c. Confirm that the results in parts (a) and (b) are consistent and compare the work involved in the two methods.

Short answer

In conclusion, both methods for evaluating the integral \(I=\int \sin^2 x \cos^2 x dx\) are consistent, giving us two equivalent expressions for the integral: Method 1: \(I = \frac{1}{4}x - \frac{1}{32}\sin 4x + C\) Method 2: \(I = \frac{1}{3}\sin^3 x - \frac{1}{5}\sin^5 x + C\) The first method, which utilized the identity \(\sin 2x = 2\sin x\cos x\), was more straightforward, as it led directly to an expression in terms of x. The second method required extra differentiation steps to ensure the results were consistent. Nonetheless, both methods are valid and produce correct results for the given integral.

Step-by-step solution

Solution using Identity \(\sin 2x = 2\sin x\cos x\)

First, rewrite the integrand with the identity: $$\sin^2 x \cos^2 x = \frac{1}{4} (\sin 2x)^2$$ Now, we have the integral: $$I = \int \frac{1}{4} (\sin 2x)^2 dx$$ Let \(u = 2x\), so \(\frac{1}{2}du = dx\). The integral becomes: $$I = \frac{1}{4} \int (\sin u)^2 \cdot \frac{1}{2} du$$ $$I = \frac{1}{8} \int (\sin u)^2 du$$ Now, use the identity \(\cos 2u = 1 - 2\sin^2 u\): $$\sin^2 u = \frac{1-\cos 2u}{2}$$ The integral becomes: $$I = \frac{1}{8} \int \left(\frac{1-\cos 2u}{2}\right) du$$ Integrate this expression: $$I = \frac{1}{8} \left[\frac{1}{2}u - \frac{1}{4}\sin 2u \right] + C$$ Substitute \(u=2x\) back into the expression: $$I = \frac{1}{8} \left[\frac{1}{2}(2x) - \frac{1}{4}\sin(2(2x)) \right] + C$$ $$I = \frac{1}{4}x - \frac{1}{32}\sin 4x + C$$

Solution using Identity \(\cos^2 x = 1-\sin^2 x\)

Rewrite the integrand using the identity: $$\sin^2 x \cos^2 x = \sin^2 x(1-\sin^2 x)$$ Integrate this expression: $$I = \int \sin^2 x(1-\sin^2 x) dx$$ Let \(v = \sin x\), so \(\cos x\, dv = dx\). The integral becomes: $$I = \int v^2 (1-v^2) dv$$ Now, expand the integrand: $$I = \int (v^2 - v^4) dv$$ Integrate this expression: $$I = \left[\frac{1}{3}v^3 - \frac{1}{5}v^5\right] + C$$ Substitute \(v=\sin x\) back into the expression: $$I = \frac{1}{3}\sin^3 x - \frac{1}{5}\sin^5 x + C$$

Comparing the two methods and checking consistency

We have obtained two different expressions for the integral \(I\) using the two methods: Method 1: \(I = \frac{1}{4}x - \frac{1}{32}\sin 4x + C\) Method 2: \(I = \frac{1}{3}\sin^3 x - \frac{1}{5}\sin^5 x + C\) To check if both expressions are consistent, we need to find the derivative of the second integral expression and see if it equals the integrand. Taking the derivative, we get: $$\frac{dI}{dx} = -\cos^3 x\sin x + 5\cos x\sin^4 x$$ Now, writing the expression in terms of \(\sin x\) using the identity \(\cos^2 x = 1 - \sin^2 x\): $$\frac{dI}{dx} = \sin x\left[-(1-\sin^2 x)^\frac{3}{2} + 5\sin^3 x\right]$$ $$\frac{dI}{dx} = \sin^2 x\cos^2 x$$ We can see that the derivative of the expression from method 2 matches the integrand, which confirms that both methods are consistent. Comparing the two methods, it can be seen that the first method using the identity \(\sin 2x = 2\sin x\cos x\) was more straightforward as it directly gave an expression in terms of x, while the second method required extra differentiation steps to make sure the results are consistent.

Key concepts

Trigonometric Identities

The beauty of trigonometric identities lies in their ability to simplify complex expressions and facilitate the calculation of integrals and derivatives in calculus. A fundamental understanding of identities like \(\sin^2 x + \cos^2 x = 1\), \(\sin 2x = 2\sin x\cos x\), and \(\cos 2x = 1 - 2\sin^2 x\) – among others – is essential. Trigonometric identities transform the integrals into a more manageable form, which is especially helpful in integrating products of trigonometric functions, such as in the given exercise \(I=\int \sin ^{2} x \cos ^{2} x d x\).

By employing the identities \(\sin 2x = 2\sin x\cos x\) or \(\cos ^{2} x = 1-\sin ^{2} x\), students can rewrite the integrand to avoid dealing directly with the product of \(\sin x\) and \(\cos x\) raised to powers. This makes the integral more tractable and highlights the power of trigonometric identities in simplifying integrals. Not only do these identities enable the integration process, but they also reveal the interconnectedness of trigonometric functions.

Integration Techniques

Integration techniques encompass a variety of strategies used to find the antiderivative of functions. The goal is to express the integral in a basic form that can be easily recognized and evaluated. In the context of integrating trigonometric functions, these techniques often require the use of trigonometric identities, substitution, and algebraic manipulation.

For instance, when integrating \(\sin^{2}x \cos^{2}x\), employing identities and making substitutions are two pivotal steps. The identity \(\sin 2x = 2\sin x\cos x\) can be used to convert the integrand into a form that involves \(\sin^2(2x)\), while a substitution such as \(u = 2x\) simplifies the integral further. Another technique involves using the identity \(\cos^{2}x = 1 - \sin^{2}x\) to separate the original integral into two simpler terms. Both techniques lead to an integral that can be solved with basic knowledge of antiderivatives, showcasing the practicality of integration techniques in calculus.

Calculus

Calculus is a branch of mathematics that deals with rates of change (differential calculus) and accumulation of quantities (integral calculus). It forms the foundation for many scientific and engineering disciplines. Understanding calculus concepts such as derivatives and integrals is vital for solving a wide variety of problems involving change and area respectively.

In the exercise \(I=\int \sin ^{2} x \cos ^{2} x d x\), calculus is used to find the integral of a trigonometric function, which in itself demonstrates the process of accumulating small amounts of area under a curve. In both methods outlined, calculus not only allows us to evaluate the integral but also to confirm the results through differentiation—a key aspect of differential calculus. By taking the derivative of the obtained integrals and checking they match the original integrand, one can ensure the accuracy and consistency of the integration methods. This process exemplifies the deep relationship between the principal operations of calculus: differentiation and integration.