Related Formula
King's Rule Variant: ∫₋ₐa f(x) dx = ∫₀a (f(x) + f(-x)) dx$$\text{King's Rule Variant: } \int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$$
Core Logic
Let the integral be I$I$.
We apply the property ∫₋ₐ^a f(x) dx = ∫₀^a (f(x)+f(-x)) dx$\int_{-a}^a f(x) dx = \int_0^a (f(x)+f(-x)) dx$ to each fractional term independently.
For the first term f₁(x) = (x² x)/(1+π^x)$f_1(x) = \frac{x^2\cos x}{1+\pi^x}$:
f₁(x) + f₁(-x) = (x² x)/(1+π^x) + (-x)² (-x)1+π-x = x² x ( (1)/(1+π^x) + (π^x)/(π^x+1) ) = x² x$$f_1(x) + f_1(-x) = \frac{x^2\cos x}{1+\pi^x} + \frac{(-x)^2\cos(-x)}{1+\pi^{-x}} = x^2\cos x \left( \frac{1}{1+\pi^x} + \frac{\pi^x}{\pi^x+1} \right) = x^2\cos x$$
For the second term f₂(x) = 1+ ²x1+e^ x²⁰²³$f_2(x) = \frac{1+\sin^2x}{1+e^{\sin x^{2023}}}$:
Observe the exponent: (-x)²⁰²³ = (-x²⁰²³) = - x²⁰²³$\sin(-x)^{2023} = \sin(-x^{2023}) = -\sin x^{2023}$.
f₂(x) + f₂(-x) = 1+ ²x1+ep + 1+ ²(-x)1+e-p = (1+ ²x) ( (1)/(1+e^p) + (e^p)/(e^p+1) ) = 1+ ²x$$f_2(x) + f_2(-x) = \frac{1+\sin^2x}{1+e^{p}} + \frac{1+\sin^2(-x)}{1+e^{-p}} = (1+\sin^2x) \left( \frac{1}{1+e^p} + \frac{e^p}{e^p+1} \right) = 1+\sin^2x$$
Thus, the integral radically simplifies to:
I = ∫₀π/2 (x² x + 1 + ² x) dx$$I = \int_0^{\pi/2} (x^2\cos x + 1 + \sin^2 x) dx$$
Step 1: Execute Integration by Parts
Evaluate ∫₀π/2 x² x dx$\int_0^{\pi/2} x^2\cos x \, dx$ using integration by parts:
u = x² ⇒ du = 2x dx$$u = x^2 \Rightarrow du = 2x \, dx$$
dv = x dx ⇒ v = x$$dv = \cos x \, dx \Rightarrow v = \sin x$$
∫ x² x dx = x² x - ∫ 2x x dx$$\int x^2\cos x dx = x^2\sin x - \int 2x\sin x dx$$
Applying parts again to ∫ 2x x dx$\int 2x\sin x dx$:
∫ 2x x dx = 2x(- x) - ∫ 2(- x)dx = -2x x + 2 x$$\int 2x\sin x dx = 2x(-\cos x) - \int 2(-\cos x)dx = -2x\cos x + 2\sin x$$
Substituting back:
[ x² x + 2x x - 2 x ]₀π/2$$\left[ x^2\sin x + 2x\cos x - 2\sin x \right]_0^{\pi/2}$$
Evaluate at upper limit π/2$\pi/2$:
(π/2)²(1) + 0 - 2(1) = (π²)/(4) - 2$$(\pi/2)^2(1) + 0 - 2(1) = \frac{\pi^2}{4} - 2$$
Evaluate at lower limit 0:
0 + 0 - 0 = 0$0 + 0 - 0 = 0$
So, ∫₀π/2 x² x dx = (π²)/(4) - 2$\int_0^{\pi/2} x^2\cos x \, dx = \frac{\pi^2}{4} - 2$.
Step 2: Execute Standard Integrals
Evaluate ∫₀π/2 (1 + ² x) dx$\int_0^{\pi/2} (1 + \sin^2 x) dx$:
= ∫₀π/2 dx + ∫₀π/2 ² x dx$$= \int_0^{\pi/2} dx + \int_0^{\pi/2} \sin^2 x dx$$
= (π)/(2) + [ (1)/(2) · (π)/(2) ] = (π)/(2) + (π)/(4) = (3π)/(4)$$= \frac{\pi}{2} + \left[ \frac{1}{2} \cdot \frac{\pi}{2} \right] = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$$
Step 3: Combine and Compare
Sum the components to get the total integral I$I$:
I = ((π²)/(4) - 2) + (3π)/(4) = (π²)/(4) + (3π)/(4) - 2$$I = \left(\frac{\pi^2}{4} - 2\right) + \frac{3\pi}{4} = \frac{\pi^2}{4} + \frac{3\pi}{4} - 2$$
Factor out (π)/(4)$\frac{\pi}{4}$:
I = (π)/(4)(π + 3) - 2$$I = \frac{\pi}{4}(\pi + 3) - 2$$
The problem states I = (π)/(4)(π + a) - 2$I = \frac{\pi}{4}(\pi + a) - 2$. Comparing the two expressions gives:
a = 3$a = 3$
Pattern Recognition
Frightening denominators like 1+af(x)$1+a^{f(x)}$ in an integral from -L$-L$ to L$L$ where f(x)$f(x)$ is an odd function are almost exclusively designed to cancel out and leave 1 via the f(x)+f(-x)$f(x)+f(-x)$ expansion. Strip the ugly denominators immediately.
Chapter Mix
Class 12 Mathematics: Integral Calculus