Understanding the Problem: Definite Integrals and Periodic Functions

The problem asks us to find the value of a definite integral over a larger interval, given the value of the same integral over a smaller interval. Both integrals involve the function \(f(x) = \sin^4 x + \cos^4 x\). The key to solving this problem lies in understanding the properties of definite integrals, particularly those involving periodic functions.

Identifying the Function and its Periodicity

The function inside the integral is \(f(x) = \sin^4 x + \cos^4 x\). Let's try to simplify this function and determine its period.

We can rewrite the function using the identity \(\sin^2 x + \cos^2 x = 1\):

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

\(f(x) = (1)^2 - 2 (\sin x \cos x)^2\)

\(f(x) = 1 - 2 \left(\frac{1}{2} \sin (2x)\right)^2\)

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

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

Now, let's determine the period of \(f(x)\). The period of \(\sin(2x)\) is \(\frac{2\pi}{2} = \pi\). The period of \(\sin^2(u)\) is \(\pi\), so the period of \(\sin^2(2x)\) is \(\frac{\pi}{2}\).

Let's formally check if \(\pi/2\) is the period:

\(f(x + \pi/2) = \sin^4 (x + \pi/2) + \cos^4 (x + \pi/2)\)

Using trigonometric identities, \(\sin (x + \pi/2) = \cos x\) and \(\cos (x + \pi/2) = -\sin x\).

\(f(x + \pi/2) = (\cos x)^4 + (-\sin x)^4\)

\(f(x + \pi/2) = \cos^4 x + \sin^4 x\)

\(f(x + \pi/2) = f(x)\)

Thus, the function \(f(x) = \sin^4 x + \cos^4 x\) is periodic with a period \(T = \pi/2\).

Property of Definite Integrals for Periodic Functions

For a periodic function \(f(x)\) with period \(T\), the definite integral over an interval of length \(nT\) is \(n\) times the integral over one period \(T\), provided the interval starts from 0. Mathematically:

\(\displaystyle\int_0^{nT} f(x) dx = n \int_0^T f(x) dx\)

Applying the Property to Solve the Problem

We are given the integral \(\displaystyle\int_0^{\pi/2} (\sin^4 x + \cos^4 x) dx = k\).

Since the period of \(f(x) = \sin^4 x + \cos^4 x\) is \(T = \pi/2\), the given integral \(\displaystyle\int_0^{\pi/2} f(x) dx\) is the integral over exactly one period starting from 0.

So, \(\displaystyle\int_0^T f(x) dx = k\).

We need to find the value of \(\displaystyle\int_0^{20 \pi} (\sin^4 x + \cos^4 x) dx\).

The upper limit of the integral is \(20 \pi\). We need to express this limit in terms of the period \(T = \pi/2\).

Let \(20 \pi = nT = n (\pi/2)\).

Solving for \(n\): \(n = \frac{20 \pi}{\pi/2} = 20 \pi \times \frac{2}{\pi} = 40\).

So the integral is from 0 to \(40 T\).

Using the property \(\displaystyle\int_0^{nT} f(x) dx = n \int_0^T f(x) dx\) with \(n = 40\) and \(T = \pi/2\):

\(\displaystyle\int_0^{20 \pi} (\sin^4 x + \cos^4 x) dx = \int_0^{40 (\pi/2)} f(x) dx = 40 \int_0^{\pi/2} f(x) dx\)

We know that \(\displaystyle\int_0^{\pi/2} f(x) dx = k\).

Therefore, \(\displaystyle\int_0^{20 \pi} (\sin^4 x + \cos^4 x) dx = 40 \times k = 40k\).

Final Answer

The value of the integral \(\displaystyle\int_0^{20 \pi}\) (sin 4 x + cos 4  x)dx is \(40k\).

Revision Table: Key Concepts

Additional Information: Integral of \(\sin^4 x + \cos^4 x\)

While not strictly necessary to solve this specific problem (since \(k\) is given), we can actually evaluate the integral \(\displaystyle\int_0^{\pi/2} (\sin^4 x + \cos^4 x) dx\) directly.

We know \(f(x) = 1 - \frac{1}{2} \sin^2 (2x)\).

Using the identity \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\):

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

\(f(x) = 1 - \frac{1}{4} (1 - \cos(4x))\)

\(f(x) = 1 - \frac{1}{4} + \frac{1}{4} \cos(4x)\)

\(f(x) = \frac{3}{4} + \frac{1}{4} \cos(4x)\)

Now, integrate from 0 to \(\pi/2\):

\(\displaystyle\int_0^{\pi/2} \left(\frac{3}{4} + \frac{1}{4} \cos(4x)\right) dx\)

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

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

Evaluate at the limits:

At \(x = \pi/2\): \(\frac{3}{4} \left(\frac{\pi}{2}\right) + \frac{1}{16} \sin(4 \times \pi/2) = \frac{3\pi}{8} + \frac{1}{16} \sin(2\pi) = \frac{3\pi}{8} + \frac{1}{16} (0) = \frac{3\pi}{8}\)

At \(x = 0\): \(\frac{3}{4} (0) + \frac{1}{16} \sin(0) = 0 + 0 = 0\)

So, \(\displaystyle\int_0^{\pi/2} (\sin^4 x + \cos^4 x) dx = \frac{3\pi}{8}\).

This means \(k = \frac{3\pi}{8}\).

Then the value of the second integral is \(40k = 40 \times \frac{3\pi}{8} = 5 \times 3\pi = 15\pi\).

This confirms that the approach using periodicity is correct and gives the answer in terms of \(k\).