Evaluating the Definite Integral $\int_0^\pi (\sin^4 x + \cos^4 x) dx$

The question asks us to find the value of the definite integral $\int_0^\pi (\sin^4 x + \cos^4 x) dx$. This requires us to first simplify the integrand $\sin^4 x + \cos^4 x$ and then perform the integration over the given limits.

Simplifying the Integrand $\sin^4 x + \cos^4 x$ using Trigonometric Identities

We can simplify the expression $\sin^4 x + \cos^4 x$ using the fundamental identity $\sin^2 x + \cos^2 x = 1$. Let's square this identity:

$\displaystyle (\sin^2 x + \cos^2 x)^2 = 1^2$

Expanding the left side, we get:

$\displaystyle \sin^4 x + 2\sin^2 x \cos^2 x + \cos^4 x = 1$

Rearranging the terms to isolate $\sin^4 x + \cos^4 x$:

$\displaystyle \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$

Now, we use the double angle identity for sine, which is $\sin 2x = 2 \sin x \cos x$. Squaring both sides gives $\sin^2 2x = 4 \sin^2 x \cos^2 x$. Therefore, $2 \sin^2 x \cos^2 x = \frac{1}{2} \sin^2 2x$. Substituting this back into the expression:

$\displaystyle \sin^4 x + \cos^4 x = 1 - \frac{1}{2} \sin^2 2x$

Next, we use the identity $\cos 2\theta = 1 - 2 \sin^2 \theta$, which can be rearranged to $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$. Applying this to $\sin^2 2x$ (with $\theta = 2x$), we get:

$\displaystyle \sin^2 2x = \frac{1 - \cos(2 \cdot 2x)}{2} = \frac{1 - \cos 4x}{2}$

Substitute this back into the expression for $\sin^4 x + \cos^4 x$:

$\displaystyle \sin^4 x + \cos^4 x = 1 - \frac{1}{2} \left(\frac{1 - \cos 4x}{2}\right)$

$\displaystyle \sin^4 x + \cos^4 x = 1 - \frac{1 - \cos 4x}{4}$

$\displaystyle \sin^4 x + \cos^4 x = \frac{4 - (1 - \cos 4x)}{4}$

$\displaystyle \sin^4 x + \cos^4 x = \frac{4 - 1 + \cos 4x}{4}$

$\displaystyle \sin^4 x + \cos^4 x = \frac{3 + \cos 4x}{4}$

So, the integrand simplifies to $\frac{3}{4} + \frac{1}{4} \cos 4x$.

Calculating the Definite Integral $\int_0^\pi (\frac{3}{4} + \frac{1}{4} \cos 4x) dx$

Now, we can evaluate the definite integral:

$\displaystyle \int_0^\pi (\sin^4 x + \cos^4 x) dx = \int_0^\pi \left(\frac{3}{4} + \frac{1}{4} \cos 4x\right) dx$

We can integrate term by term:

$\displaystyle \int \frac{3}{4} dx = \frac{3}{4} x$

$\displaystyle \int \frac{1}{4} \cos 4x dx = \frac{1}{4} \int \cos 4x dx$

Let $u = 4x$, then $du = 4 dx$, so $dx = \frac{1}{4} du$.

$\displaystyle \frac{1}{4} \int \cos u \left(\frac{1}{4} du\right) = \frac{1}{16} \int \cos u du = \frac{1}{16} \sin u + C = \frac{1}{16} \sin 4x + C$

So, the indefinite integral is $\frac{3}{4} x + \frac{1}{16} \sin 4x$. Now we apply the limits of integration from $0$ to $\pi$:

$\displaystyle \left[\frac{3}{4} x + \frac{1}{16} \sin 4x\right]_0^\pi$

Evaluate the expression at the upper limit $x = \pi$:

$\displaystyle \left(\frac{3}{4} \pi + \frac{1}{16} \sin (4\pi)\right)$

Since $\sin(4\pi) = 0$, this becomes:

$\displaystyle \frac{3}{4} \pi + \frac{1}{16} (0) = \frac{3}{4} \pi$

Evaluate the expression at the lower limit $x = 0$:

$\displaystyle \left(\frac{3}{4} (0) + \frac{1}{16} \sin (4 \cdot 0)\right)$

Since $\sin(0) = 0$, this becomes:

$\displaystyle 0 + \frac{1}{16} (0) = 0$

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

$\displaystyle \int_0^\pi (\sin^4 x + \cos^4 x) dx = \frac{3}{4} \pi - 0 = \frac{3\pi}{4}$

Thus, the value of the definite integral is $\frac{3\pi}{4}$.

Step-by-Step Solution Summary for Definite Integral

Here is a summary of the steps taken to evaluate the definite integral:

1. Simplify the integrand $\sin^4 x + \cos^4 x$ using trigonometric identities: $\sin^4 x + \cos^4 x = \frac{3}{4} + \frac{1}{4} \cos 4x$.
2. Integrate the simplified expression: $\int \left(\frac{3}{4} + \frac{1}{4} \cos 4x\right) dx = \frac{3}{4} x + \frac{1}{16} \sin 4x + C$.
3. Apply the limits of integration from $0$ to $\pi$: $\left[\frac{3}{4} x + \frac{1}{16} \sin 4x\right]_0^\pi$.
4. Evaluate at the upper limit $x=\pi$: $\frac{3}{4} \pi + \frac{1}{16} \sin(4\pi) = \frac{3}{4} \pi$.
5. Evaluate at the lower limit $x=0$: $\frac{3}{4} (0) + \frac{1}{16} \sin(0) = 0$.
6. Subtract the lower limit value from the upper limit value: $\frac{3}{4} \pi - 0 = \frac{3\pi}{4}$.

Integral Evaluation Revision

Let's review the key aspects of evaluating this definite integral.

• The first crucial step was simplifying the powers of sine and cosine using fundamental identities and double angle formulas. This transformed the integral into a simpler form involving $\cos 4x$.
• The second step involved basic integration rules for constant terms and cosine functions. Remember that $\int \cos(ax) dx = \frac{1}{a} \sin(ax) + C$.
• Finally, applying the limits of integration involves evaluating the antiderivative at the upper limit and subtracting its value at the lower limit. Special attention must be paid to trigonometric function values at the limits (e.g., $\sin(0)$, $\sin(\pi)$, $\sin(2\pi)$, etc.). In this case, $\sin(4\pi) = \sin(0) = 0$.

Related Integration Concepts

Understanding definite integrals and trigonometric identities is key to solving problems like this. Here are some related concepts:

• Definite Integral: Represents the area under the curve of a function between two specified limits. It is calculated using the Fundamental Theorem of Calculus.
• Fundamental Theorem of Calculus: States that if $F(x)$ is an antiderivative of $f(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$.
• Trigonometric Identities: Equations involving trigonometric functions that are true for all values of the variables where the functions are defined. Key identities used here include $\sin^2 x + \cos^2 x = 1$, $\sin 2x = 2 \sin x \cos x$, and $\cos 2x = 1 - 2 \sin^2 x$.
• Power Reduction Formulas: Identities that express powers of trigonometric functions (like $\sin^2 x$ or $\cos^2 x$) in terms of trigonometric functions raised to the first power (e.g., $\cos 2x$). These are often derived from double angle formulas.

Additional Information on Integral Properties

While not directly needed for this specific integral $\int_0^\pi (\sin^4 x + \cos^4 x) dx$, the initial context involving the integral $I=\int_{-2 \pi}^{2 \pi} \frac{\sin ^4 x+\cos ^4 x}{1+3^x} d x$ often suggests using properties of definite integrals, particularly for even/odd functions or integrals of the form $\int_{-a}^a f(x) dx$ or $\int_0^{2a} f(x) dx$. For instance, the property $\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$ is useful when the integrand has a mix of even and odd parts or a denominator like $1+a^x$. However, the question specifically asks for $\int_0^\pi (\sin^4 x + \cos^4 x) dx$, which is a standard evaluation problem once the integrand is simplified.