Question

The value of the integral \(\int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}\)

• A. \( \frac{\sqrt{2} \pi^2}{8} \)
• B. \( \frac{\sqrt{2} \pi^2}{16} \)
• C. \( \frac{\pi^2}{16\sqrt{2}} \)
• D. \( \frac{\sqrt{2} \pi^2}{64} \)

Answer 1

The Correct Option is C

\( I = \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}\)
Let \(2x = t\), then \(dx = \frac{1}{2} dt\), \[ I = \frac{1}{4} \int_{0}^{\pi} \frac{t \, dt}{\sin^4 t + \cos^4 t} \] Using symmetry: \[ I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{t + \frac{\pi}{2} - t}{\sin^4 t + \cos^4 t} \, dt \] \[ I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2}}{\sin^4 t + \cos^4 t} \, dt - I \] \[ 2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t} \] Let \(\tan t = y\), then \(\sec^2 t \, dt = dy\): \[ 2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{(1 + y^2) \, dy}{1 + y^4} \] \[ 2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{dy}{y^2 + 1} \] Let \(y = p\), then: \[ I = \frac{\pi}{16} \int_{0}^{\infty} \frac{dp}{p^2 + (\sqrt{2})^2} \] Using the standard integral formula: \[ I = \frac{\pi}{16 \sqrt{2}} \left[ \tan^{-1} \left( \frac{p}{\sqrt{2}} \right) \right]_{0}^{\infty} \] \[ I = \frac{\pi}{16 \sqrt{2}} \cdot \frac{\pi}{2} \] \[ I = \frac{\pi^2}{16 \sqrt{2}} \]

Answer 2

We are asked to evaluate the definite integral \( I = \int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)} \).

Concept Used:

To solve this integral, we will use the following concepts:

1. King's Property of Definite Integrals: For an integral of the form \( \int_0^a f(x) \, dx \), we have the property:

\[ \int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx \]

This property is particularly useful for integrals containing an \( x \) term in the numerator multiplied by a symmetric function.

2. Trigonometric Identities: The denominator can be simplified using the identity:

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

3. Standard Integration Formula:

\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \]

Step-by-Step Solution:

Step 1: Apply the King's Property to the integral.

Let the given integral be \(I\).

\[ I = \int_0^{\frac{\pi}{4}} \frac{x}{\sin^4(2x) + \cos^4(2x)} \, dx \quad \cdots(1) \]

Using the property \( \int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx \) with \( a = \frac{\pi}{4} \):

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

The terms in the denominator become:

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

So the denominator is \( \cos^4(2x) + \sin^4(2x) \), which is the same as the original. Thus, the integral becomes:

\[ I = \int_0^{\frac{\pi}{4}} \frac{\frac{\pi}{4} - x}{\sin^4(2x) + \cos^4(2x)} \, dx \quad \cdots(2) \]

Step 2: Add equations (1) and (2) to eliminate \( x \) from the numerator.

\[ 2I = \int_0^{\frac{\pi}{4}} \frac{x + (\frac{\pi}{4} - x)}{\sin^4(2x) + \cos^4(2x)} \, dx \] \[ 2I = \int_0^{\frac{\pi}{4}} \frac{\frac{\pi}{4}}{\sin^4(2x) + \cos^4(2x)} \, dx \] \[ 2I = \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{1}{\sin^4(2x) + \cos^4(2x)} \, dx \]

Step 3: Simplify the denominator and evaluate the new integral.

Using the identity \( \sin^4\theta + \cos^4\theta = 1 - \frac{1}{2}\sin^2(2\theta) \), with \( \theta = 2x \):

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

Substituting this into the integral for \(2I\):

\[ 2I = \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{1}{1 - \frac{1}{2}\sin^2(4x)} \, dx = \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{2}{2 - \sin^2(4x)} \, dx \]

Let \( u = 4x \), so \( du = 4 \, dx \) or \( dx = \frac{du}{4} \). The limits of integration change from \( x=0 \to u=0 \) and \( x=\frac{\pi}{4} \to u=\pi \).

\[ 2I = \frac{\pi}{2} \int_0^{\pi} \frac{1}{2 - \sin^2(u)} \frac{du}{4} = \frac{\pi}{8} \int_0^{\pi} \frac{1}{2 - \sin^2(u)} \, du \]

The integrand \( f(u) = \frac{1}{2 - \sin^2(u)} \) is symmetric about \( u = \frac{\pi}{2} \) (i.e., \( f(\pi - u) = f(u) \)). So, \( \int_0^{\pi} f(u) \, du = 2 \int_0^{\frac{\pi}{2}} f(u) \, du \).

\[ 2I = \frac{\pi}{8} \cdot 2 \int_0^{\frac{\pi}{2}} \frac{1}{2 - \sin^2(u)} \, du = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{1}{2 - \sin^2(u)} \, du \]

Divide the numerator and denominator by \( \cos^2(u) \):

\[ 2I = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\sec^2(u)}{2\sec^2(u) - \tan^2(u)} \, du = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\sec^2(u)}{2(1 + \tan^2(u)) - \tan^2(u)} \, du \] \[ 2I = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\sec^2(u)}{2 + \tan^2(u)} \, du \]

Step 4: Perform a second substitution to solve the integral.

Let \( v = \tan(u) \), so \( dv = \sec^2(u) \, du \). The limits change from \( u=0 \to v=0 \) and \( u=\frac{\pi}{2} \to v=\infty \).

\[ 2I = \frac{\pi}{4} \int_0^{\infty} \frac{1}{2 + v^2} \, dv = \frac{\pi}{4} \int_0^{\infty} \frac{1}{(\sqrt{2})^2 + v^2} \, dv \]

Using the standard integral \( \int \frac{1}{a^2 + x^2} dx = \frac{1}{a}\tan^{-1}(\frac{x}{a}) \):

\[ 2I = \frac{\pi}{4} \left[ \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{v}{\sqrt{2}}\right) \right]_0^{\infty} \]

Final Computation & Result:

Substitute the limits of integration:

\[ 2I = \frac{\pi}{4\sqrt{2}} \left( \lim_{v \to \infty} \tan^{-1}\left(\frac{v}{\sqrt{2}}\right) - \tan^{-1}(0) \right) \] \[ 2I = \frac{\pi}{4\sqrt{2}} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi^2}{8\sqrt{2}} \]

Now, solve for \(I\):

\[ I = \frac{\pi^2}{16\sqrt{2}} \]

The value of the integral is \( \frac{\pi^2}{16\sqrt{2}} \).