Related Formula
King's Property of Definite Integrals:
∫ₐb f(x) dx = ∫ₐb f(a+b-x) dx$$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$$
Core Logic
Let the given integral be I$I$:
I = ∫₀(π)/(4) x dx ⁴(2x)+ ⁴(2x)$$I = \int_{0}^{\frac{\pi}{4}}\frac{x \, dx}{\sin^{4}(2x)+\cos^{4}(2x)}$$
Substitute 2x = t 2dx = dt dx = (1)/(2)dt$2x = t \implies 2dx = dt \implies dx = \frac{1}{2}dt$.
When x = 0 t = 0$x = 0 \implies t = 0$, and when x = (π)/(4) t = (π)/(2)$x = \frac{\pi}{4} \implies t = \frac{\pi}{2}$.
I = (1)/(4) ∫₀(π)/(2) t dt ⁴t + ⁴t (1)$$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{t \, dt}{\sin^{4}t + \cos^{4}t} \quad \implies (1) $$
Step 1: Apply King's Property
Applying the property ∫₀a f(t) dt = ∫₀a f(a-t) dt$\int_{0}^{a} f(t) \, dt = \int_{0}^{a} f(a-t) \, dt$:
I = (1)/(4) ∫₀(π)/(2) ((π)/(2) - t) dt ⁴((π)/(2)-t) + ⁴((π)/(2)-t)$$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - t\right) dt}{\sin^{4}\left(\frac{\pi}{2}-t\right) + \cos^{4}\left(\frac{\pi}{2}-t\right)}$$
I = (1)/(4) ∫₀(π)/(2) ((π)/(2) - t) dt ⁴t + ⁴t (2)$$I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - t\right) dt}{\cos^{4}t + \sin^{4}t} \quad \implies (2) $$
Adding equations (1) and (2):
2I = (1)/(4) ∫₀(π)/(2) (π)/(2) dt ⁴t + ⁴t$$2I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} dt}{\sin^{4}t + \cos^{4}t}$$
2I = (π)/(8) ∫₀(π)/(2) dt ⁴t + ⁴t$$2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sin^{4}t + \cos^{4}t} $$
2I = (π)/(8) ∫₀(π)/(2) ⁴t dt ⁴t + 1$$2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{\sec^{4}t \, dt}{\tan^{4}t + 1} $$
2I = (π)/(8) ∫₀(π)/(2) (1 + ²t) ²t dt ⁴t + 1$$2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{(1 + \tan^{2}t)\sec^{2}t \, dt}{\tan^{4}t + 1}$$
Step 2: Substitution and Algebraic Limits
Let t = y ²t dt = dy$\tan t = y \implies \sec^{2}t \, dt = dy$.
When t = 0 y = 0$t = 0 \implies y = 0$, and when t = (π)/(2) y = ∞$t = \frac{\pi}{2} \implies y = \infty$.
2I = (π)/(8) ∫₀∞ (1 + y²) dy1 + y⁴$$2I = \frac{\pi}{8} \int_{0}^{\infty} \frac{(1 + y^{2}) \, dy}{1 + y^{4}} $$
I = (π)/(16) ∫₀∞ 1 + 1y²y² + 1y² dy$$I = \frac{\pi}{16} \int_{0}^{\infty} \frac{1 + \frac{1}{y^{2}}}{y^{2} + \frac{1}{y^{2}}} \, dy $$
Now put y - (1)/(y) = p (1 + 1y²) dy = dp$y - \frac{1}{y} = p \implies \left(1 + \frac{1}{y^{2}}\right) dy = dp$.
When y → 0^+ p → -∞$y \to 0^+ \implies p \to -\infty$, and when y → ∞ p → ∞$y \to \infty \implies p \to \infty$.
Also, y² + 1y² = p² + 2 = p² + (√(2))²$y^{2} + \frac{1}{y^{2}} = p^{2} + 2 = p^{2} + (\sqrt{2})^{2}$.
I = (π)/(16) ∫-∞∞ dpp² + (√(2))²$$I = \frac{\pi}{16} \int_{-\infty}^{\infty} \frac{dp}{p^{2} + (\sqrt{2})^{2}} $$
I = π16√(2) [ ⁻¹( p√(2)) ]-∞∞$$I = \frac{\pi}{16\sqrt{2}} \left[ \tan^{-1}\left(\frac{p}{\sqrt{2}}\right) \right]_{-\infty}^{\infty} $$
I = π16√(2) ( (π)/(2) - (-(π)/(2)) ) = π²16√(2) = √(2)π²32$$I = \frac{\pi}{16\sqrt{2}} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = \frac{\pi^{2}}{16\sqrt{2}} = \frac{\sqrt{2}\pi^{2}}{32} $$$$
Pattern Recognition
Sees: Integrand containing
$
Pattern Recognition
Sees: Integrand containing $
x
in the numerator and symmetric trigonometric functions in the denominator. Shortcut: The elimination of$ in the numerator and symmetric trigonometric functions in the denominator.
Shortcut: The elimination of $x
using King's property is standard. For integrals containing$ using King's property is standard. For integrals containing $\frac{1+y^2}{1+y^4}
, dividing by$, dividing by $y^2
transforms the denominator into a perfect square form$ transforms the denominator into a perfect square form $\left(y-\frac{1}{y}\right)^2+2$, making substitution trivial.
Chapter Mix
Class 12 Mathematics: Definite Integrals
Class 11 Mathematics: Trigonometric Identities
