Understanding the Definite Integral Problem

The problem asks us to evaluate a specific definite integral:

\[ I=\int_{-2 \pi}^{2 \pi} \frac{\sin ^4 x+\cos ^4 x}{1+3^x} d x \]

This is a definite integral with symmetric limits, ranging from \(-2\pi\) to \(2\pi\). Integrals with symmetric limits often utilize properties that simplify the integrand, especially when the integrand has a specific form like the one involving \(1+a^x\) in the denominator.

Applying Integral Properties for Symmetric Limits

For a definite integral \(\int_{-a}^{a} f(x) dx\), we can use the property:

\[ \int_{-a}^{a} f(x) dx = \int_0^{a} [f(x) + f(-x)] dx \]

In our case, \(a = 2\pi\) and \(f(x) = \frac{\sin^4 x + \cos^4 x}{1+3^x}\).

Let's find \(f(-x)\):

\[ f(-x) = \frac{\sin^4 (-x) + \cos^4 (-x)}{1+3^{-x}} \]

Since \(\sin(-x) = -\sin x\) and \(\cos(-x) = \cos x\), we have \(\sin^4 (-x) = (-\sin x)^4 = \sin^4 x\) and \(\cos^4 (-x) = (\cos x)^4 = \cos^4 x\). Thus,

\[ f(-x) = \frac{\sin^4 x + \cos^4 x}{1+3^{-x}} \]

Simplifying the Integrand \(f(x) + f(-x)\)

Now we compute \(f(x) + f(-x)\):

\[ f(x) + f(-x) = \frac{\sin^4 x + \cos^4 x}{1+3^x} + \frac{\sin^4 x + \cos^4 x}{1+3^{-x}} \]

We can factor out the numerator \(\sin^4 x + \cos^4 x\):

\[ f(x) + f(-x) = (\sin^4 x + \cos^4 x) \left( \frac{1}{1+3^x} + \frac{1}{1+3^{-x}} \right) \]

Let's simplify the term in the parenthesis:

\[ \frac{1}{1+3^x} + \frac{1}{1+3^{-x}} = \frac{1}{1+3^x} + \frac{1}{1+\frac{1}{3^x}} \]

\[ = \frac{1}{1+3^x} + \frac{1}{\frac{3^x+1}{3^x}} = \frac{1}{1+3^x} + \frac{3^x}{3^x+1} \]

\[ = \frac{1+3^x}{1+3^x} = 1 \]

Therefore, \(f(x) + f(-x) = \sin^4 x + \cos^4 x\).

The definite integral simplifies to:

\[ I = \int_0^{2 \pi} (\sin^4 x + \cos^4 x) dx \]

Simplifying the Trigonometric Expression \(\sin^4 x + \cos^4 x\)

We can rewrite \(\sin^4 x + \cos^4 x\) using algebraic and trigonometric identities:

\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x \]

Using the identity \(\sin^2 x + \cos^2 x = 1\):

\[ = (1)^2 - 2 (\sin x \cos x)^2 = 1 - 2 \left(\frac{1}{2} \sin(2x)\right)^2 \]

Using the identity \(\sin(2x) = 2 \sin x \cos x\):

\[ = 1 - 2 \left(\frac{1}{4} \sin^2(2x)\right) = 1 - \frac{1}{2} \sin^2(2x) \]

Now, using the identity \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\) with \(\theta = 2x\):

\[ = 1 - \frac{1}{2} \left(\frac{1 - \cos(4x)}{2}\right) = 1 - \frac{1 - \cos(4x)}{4} \]

\[ = \frac{4 - (1 - \cos(4x))}{4} = \frac{4 - 1 + \cos(4x)}{4} = \frac{3 + \cos(4x)}{4} \]

So, \(\sin^4 x + \cos^4 x = \frac{3}{4} + \frac{1}{4} \cos(4x)\).

Evaluating the Simplified Definite Integral

The integral now becomes:

\[ I = \int_0^{2 \pi} \left( \frac{3}{4} + \frac{1}{4} \cos(4x) \right) dx \]

We can integrate term by term:

\[ I = \left[ \frac{3}{4} x + \frac{1}{4} \frac{\sin(4x)}{4} \right]_0^{2 \pi} \]

\[ I = \left[ \frac{3}{4} x + \frac{1}{16} \sin(4x) \right]_0^{2 \pi} \]

Now, evaluate the expression at the upper and lower limits:

At \(x = 2\pi\):

\[ \frac{3}{4}(2\pi) + \frac{1}{16} \sin(4 \cdot 2\pi) = \frac{6\pi}{4} + \frac{1}{16} \sin(8\pi) = \frac{3\pi}{2} + \frac{1}{16}(0) = \frac{3\pi}{2} \]

At \(x = 0\):

\[ \frac{3}{4}(0) + \frac{1}{16} \sin(4 \cdot 0) = 0 + \frac{1}{16} \sin(0) = 0 + 0 = 0 \]

Subtracting the value at the lower limit from the value at the upper limit:

\[ I = \frac{3\pi}{2} - 0 = \frac{3\pi}{2} \]

Final Result

The value of the definite integral \(I\) is \(\frac{3\pi}{2}\).

Revision Table: Key Concepts for Definite Integrals

• Symmetric Limits Property: For an integral \(\int_{-a}^{a} f(x) dx\), it equals \(\int_0^a [f(x) + f(-x)] dx\). This is very useful when the integrand has terms like \(a^x\) or when \(f(x)\) is neither purely even nor odd.
• Trigonometric Identities: Mastering identities like \(\sin^2 x + \cos^2 x = 1\), \(\sin(2x) = 2 \sin x \cos x\), and \(\sin^2 x = \frac{1 - \cos(2x)}{2}\) is crucial for simplifying trigonometric expressions within integrals.
• Evaluating Definite Integrals: Once the integral is in a standard form, find the antiderivative \(F(x)\) and evaluate \(F(b) - F(a)\), where \(a\) and \(b\) are the lower and upper limits.

Additional Information: Integrals with Denominator \(1+a^x\)

Integrals of the form \(\int_{-a}^{a} \frac{g(x)}{1+b^x} dx\) can often be solved using the symmetric limits property. When applying the property, the term \(\frac{1}{1+b^x} + \frac{1}{1+b^{-x}}\) simplifies to 1. So the integral becomes \(\int_0^a g(x) dx\). This technique works when the numerator \(g(x)\) is an even function or simplifies nicely with this approach.

In this specific problem, the numerator \(\sin^4 x + \cos^4 x\) is an even function (\((\sin(-x))^4 + (\cos(-x))^4 = \sin^4 x + \cos^4 x\)).

Applying the property \(\int_{-a}^{a} f(x) dx = \int_0^a [f(x) + f(-x)] dx\) where \(f(x) = \frac{g(x)}{1+b^x}\) and \(g(x)\) is even:

\[ f(-x) = \frac{g(-x)}{1+b^{-x}} = \frac{g(x)}{1+b^{-x}} \quad (\text{since } g \text{ is even}) \]

\[ f(x) + f(-x) = \frac{g(x)}{1+b^x} + \frac{g(x)}{1+b^{-x}} = g(x) \left( \frac{1}{1+b^x} + \frac{1}{1+b^{-x}} \right) \]

\[ = g(x) \left( 1 \right) = g(x) \]

So, \(\int_{-a}^{a} \frac{g(x)}{1+b^x} dx = \int_0^a g(x) dx\) if \(g(x)\) is even.

In our problem, \(g(x) = \sin^4 x + \cos^4 x\), which is indeed an even function. Thus, our calculation \(\int_{-2 \pi}^{2 \pi} \frac{\sin ^4 x+\cos ^4 x}{1+3^x} d x = \int_0^{2 \pi} (\sin^4 x + \cos^4 x) dx\) is confirmed by this general property for even functions.