The value of the integral ∫(0 to π/4) x dx / (sin⁴(2x) + cos⁴(2x)) equals :
(1) √2π²/8
(2) √2π²/16
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Q: Wait, why do we use King's Property? Can't we just integrate directly?

That is a great question! If you try to integrate directly, you will find that the $x$ in the numerator makes it impossible to find an elementary antiderivative. King's Property is a surgical tool designed specifically for this situation. It allows us to replace $x$ with $(a-x)$, which, when added to the original integral, causes the $x$ terms to cancel out, leaving us with a much simpler integral to solve.

Q: Why divide by $\cos^4 t$ in the denominator?

This is a classic technique for integrals involving even powers of sine and cosine. By dividing by $\cos^4 t$, we transform the expression into one involving $\tan t$ and $\sec^2 t$. This is powerful because the derivative of $\tan t$ is $\sec^2 t$, which allows us to use a simple substitution to turn a trigonometric integral into a rational algebraic one.

Q: Why did the limits change to $-\infty$ to $\infty$ in the final substitution?

When we substitute $p = y - \frac{1}{y}$, we must look at the behavior of $p$ as $y$ changes. As $y$ approaches $0$ from the positive side, $\frac{1}{y}$ goes to $\infty$, so $p$ goes to $-\infty$. As $y$ approaches $\infty$, $p$ also approaches $\infty$. Thus, the limits of integration for $p$ are indeed from $-\infty$ to $\infty$.

Question

The value of the integral $\int_{0}^{\frac{\pi}{4}} \frac{x dx}{\sin^4(2x) + \cos^4(2x)}$ equals :

Options:

(A) $\frac{\sqrt{2}\pi^2}{8}$

(B) $\frac{\sqrt{2}\pi^2}{16}$

(C) $\frac{\sqrt{2}\pi^2}{32}$

(D) $\frac{\sqrt{2}\pi^2}{64}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to evaluate the integral:

$$I = \int_{0}^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}$$

It looks intimidating, but in the world of JEE Advanced, every complex problem is just a series of elegant, logical steps waiting to be uncovered. Let us embark on this journey together.

### Phase 1: Clearing the Fog

The first thing that catches our eye is the argument $2x$. To simplify our lives, we perform a substitution: let $2x = t$.

This implies $dx = \frac{1}{2} dt$. We must also update our limits of integration: when $x = 0$, $t = 0$, and when $x = \frac{\pi}{4}$, $t = \frac{\pi}{2}$.

Our integral now transforms into:

$$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{t \, dt}{\sin^4 t + \cos^4 t}$$

The fog is clearing, and the structure is becoming visible.

### Phase 2: The King's Property

Now, we face the $t$ in the numerator. It is the obstacle preventing direct integration. Here, we invoke the legendary **King's Property**:

$$\int_0^a f(t) \, dt = \int_0^a f(a-t) \, dt$$

With $a = \frac{\pi}{2}$, we replace $t$ with $(\frac{\pi}{2} - t)$. Because $\sin(\frac{\pi}{2} - t) = \cos t$ and $\cos(\frac{\pi}{2} - t) = \sin t$, the denominator remains unchanged.

The integral becomes:

$$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - t) \, dt}{\sin^4 t + \cos^4 t}$$

When we add this to our original integral, the $t$ terms vanish, leaving us with:

$$2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}$$

This is the magic of symmetry.

### Phase 3: The Tangent Transformation

We are left with a standard integral. We divide the numerator and denominator by $\cos^4 t$, yielding:

$$\int \frac{\sec^4 t \, dt}{	an^4 t + 1}$$

By substituting $	an t = y$, we get $\sec^2 t \, dt = dy$. Since $\sec^2 t = 1 + 	an^2 t = 1 + y^2$, the integral becomes:

$$2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{(1+y^2) \, dy}{1+y^4}$$

Dividing the numerator and denominator by $y^2$, we get:

$$2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{(1 + \frac{1}{y^2}) \, dy}{y^2 + \frac{1}{y^2}}$$

### Phase 4: The Final Masterstroke

Let $p = y - \frac{1}{y}$. Then $dp = (1 + \frac{1}{y^2}) \, dy$, and $y^2 + \frac{1}{y^2} = p^2 + 2$.

The limits change from $0$ to $\infty$ for $y$ to $-\infty$ to $\infty$ for $p$. The integral simplifies to:

$$2I = \frac{\pi}{8} \int_{-\infty}^{\infty} \frac{dp}{p^2 + (\sqrt{2})^2}$$

Using the standard formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})$, we evaluate this to get:

$$2I = \frac{\pi}{8} \left[ \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{p}{\sqrt{2}}ight) ight]_{-\infty}^{\infty}$$

Evaluating the limits, we get:

$$2I = \frac{\pi}{8\sqrt{2}} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}ight) ight) = \frac{\pi^2}{8\sqrt{2}}$$

Finally, $I = \frac{\pi^2}{16\sqrt{2}}$, which rationalizes to **$\frac{\sqrt{2}\pi^2}{32}$**. We have conquered the fortress!

Animated Solution for Mathematics - Definite Integration: The value of the integral $\int_{0}^{\frac{π}{4}} \frac{x d x}{s i n ^{4} ( 2 x ) + c o s ^{4} ( 2 x )}$ equals :

Select Answer:

Visualized Solution

Define the Integral $I$

Substitution $2 x = t$

Transform the Integral

Apply King's Property

Add the Two Integrals

Convert to Tangent Form

Substitution $tan t = y$

The Rational Integral Form

Substitution $y - \frac{1}{y} = p$

Integrate using $tan^{- 1}$ Formula

Evaluate Limits and Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral ∫04π​​sin4(2x)+cos4(2x)xdx​ equals :

Select Answer:

Visualized Solution

Define the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x=t

Transform the Integral

Apply King's Property

Add the Two Integrals

Convert to Tangent Form

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tant=y

The Rational Integral Form

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y−y1​=p

Integrate using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1 Formula