Calculating the Value of \(8 I_1^2\) for the Given Definite Integral

The problem asks for the value of \(8 I_1^2\), where \(I_1\) is defined by the definite integral:

\(I_1=\int_0^\pi \frac{x}{1+\cos ^2 x} d x\)

To find the value of \(8 I_1^2\), we first need to evaluate the integral \(I_1\).

Evaluating the Integral \(I_1\)

We can use the property of definite integrals which states that \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\). Here, \(a = \pi\).

Applying this property to \(I_1\):

\(I_1 = \int_0^\pi \frac{x}{1+\cos ^2 x} dx \quad \ldots (1)\)

\(I_1 = \int_0^\pi \frac{\pi-x}{1+\cos ^2 (\pi-x)} dx\)

Since \(\cos(\pi-x) = -\cos x\), we have \(\cos^2(\pi-x) = (-\cos x)^2 = \cos^2 x\). Substituting this into the equation:

\(I_1 = \int_0^\pi \frac{\pi-x}{1+\cos ^2 x} dx \quad \ldots (2)\)

Now, add equation (1) and equation (2):

\(I_1 + I_1 = \int_0^\pi \frac{x}{1+\cos ^2 x} dx + \int_0^\pi \frac{\pi-x}{1+\cos ^2 x} dx\)

\(2 I_1 = \int_0^\pi \frac{x + (\pi-x)}{1+\cos ^2 x} dx\)

\(2 I_1 = \int_0^\pi \frac{\pi}{1+\cos ^2 x} dx\)

\(I_1 = \frac{\pi}{2} \int_0^\pi \frac{1}{1+\cos ^2 x} dx\)

Evaluating the Auxiliary Integral \(\int_0^\pi \frac{1}{1+\cos ^2 x} dx\)

Let's evaluate the integral \(J = \int_0^\pi \frac{1}{1+\cos ^2 x} dx\). We can use the property \(\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx\) if \(f(2a-x) = f(x)\).

Here \(a = \pi/2\), \(2a = \pi\), and \(f(x) = \frac{1}{1+\cos ^2 x}\). We check \(f(\pi-x)\):

\(f(\pi-x) = \frac{1}{1+\cos^2(\pi-x)} = \frac{1}{1+(-\cos x)^2} = \frac{1}{1+\cos^2 x} = f(x)\)

Since \(f(\pi-x) = f(x)\), we have:

\(J = \int_0^\pi \frac{1}{1+\cos ^2 x} dx = 2 \int_0^{\pi/2} \frac{1}{1+\cos ^2 x} dx\)

Now, let's evaluate \(\int_0^{\pi/2} \frac{1}{1+\cos ^2 x} dx\). Divide the numerator and denominator by \(\cos^2 x\):

\(\int_0^{\pi/2} \frac{\frac{1}{\cos^2 x}}{\frac{1}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x}} dx = \int_0^{\pi/2} \frac{\sec^2 x}{\sec^2 x + 1} dx\)

Using the identity \(\sec^2 x = 1 + \tan^2 x\):

\(\int_0^{\pi/2} \frac{\sec^2 x}{(1 + \tan^2 x) + 1} dx = \int_0^{\pi/2} \frac{\sec^2 x}{2 + \tan^2 x} dx\)

Let \(t = \tan x\). Then \(dt = \sec^2 x dx\).

When \(x=0\), \(t = \tan 0 = 0\).

When \(x=\pi/2\), \(t = \tan(\pi/2) \to \infty\).

The integral becomes:

\(\int_0^\infty \frac{dt}{2 + t^2} = \int_0^\infty \frac{dt}{(\sqrt{2})^2 + t^2}\)

This is a standard integral of the form \(\int \frac{dx}{a^2+x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)\).

So, the value of the integral is:

\(\left[\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right)\right]_0^\infty = \frac{1}{\sqrt{2}} \left(\tan^{-1}(\infty) - \tan^{-1}(0)\right) = \frac{1}{\sqrt{2}} \left(\frac{\pi}{2} - 0\right) = \frac{\pi}{2\sqrt{2}}\)

Therefore, \(J = \int_0^\pi \frac{1}{1+\cos ^2 x} dx = 2 \times \frac{\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}\).

Finding the Value of \(I_1\)

Substitute the value of the auxiliary integral back into the expression for \(I_1\):

\(I_1 = \frac{\pi}{2} \times J = \frac{\pi}{2} \times \frac{\pi}{\sqrt{2}} = \frac{\pi^2}{2\sqrt{2}}\)

Calculating \(8 I_1^2\)

Now we need to calculate \(8 I_1^2\):

\(I_1^2 = \left(\frac{\pi^2}{2\sqrt{2}}\right)^2 = \frac{(\pi^2)^2}{(2\sqrt{2})^2} = \frac{\pi^4}{4 \times 2} = \frac{\pi^4}{8}\)

Finally, multiply by 8:

\(8 I_1^2 = 8 \times \frac{\pi^4}{8} = \pi^4\)

Conclusion

The value of \(8 I_1^2\) is \(\pi^4\).

Revision Table: Integral Properties Used

Additional Information: Trigonometric Substitutions in Integrals

Many integrals involving trigonometric functions can be solved using specific substitutions or by transforming the integrand using trigonometric identities. In this problem, we encountered \(\int \frac{\sec^2 x}{a + b\tan^2 x} dx\). This form is a strong indicator for the substitution \(t = \tan x\), because \(dt = \sec^2 x dx\), which simplifies the numerator.

Other common trigonometric substitutions include:

• For integrals involving \(\sqrt{a^2 - x^2}\), use \(x = a \sin \theta\).
• For integrals involving \(\sqrt{a^2 + x^2}\), use \(x = a \tan \theta\).
• For integrals involving \(\sqrt{x^2 - a^2}\), use \(x = a \sec \theta\).

Mastering these techniques and properties is crucial for solving definite and indefinite integrals in calculus.