Simplify the Integral using Symmetry
Let I = ∫ 0 π x 2 sin x cos x sin 4 x + cos 4 x d x I = \int_{0}^{\pi} \frac{x^{2} \sin x \cos x}{\sin ^{4} x + \cos ^{4} x} \, dx I = ∫ 0 π s i n 4 x + c o s 4 x x 2 s i n x c o s x d x ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x \int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x
I = ∫ 0 π ( π − x ) 2 sin ( π − x ) cos ( π − x ) sin 4 ( π − x ) + cos 4 ( π − x ) d x I = \int_{0}^{\pi} \frac{(\pi-x)^{2} \sin(\pi-x) \cos(\pi-x)}{\sin^4(\pi-x) + \cos^4(\pi-x)} \, dx I = ∫ 0 π sin 4 ( π − x ) + cos 4 ( π − x ) ( π − x ) 2 sin ( π − x ) cos ( π − x ) d x I = ∫ 0 π ( π 2 − 2 π x + x 2 ) sin x ( − cos x ) sin 4 x + cos 4 x d x I = \int_{0}^{\pi} \frac{(\pi^2 - 2\pi x + x^2) \sin x (-\cos x)}{\sin^4 x + \cos^4 x} \, dx I = ∫ 0 π sin 4 x + cos 4 x ( π 2 − 2 π x + x 2 ) sin x ( − cos x ) d x I = − π 2 ∫ 0 π sin x cos x sin 4 x + cos 4 x d x + 2 π ∫ 0 π x sin x cos x sin 4 x + cos 4 x d x − I I = -\pi^2 \int_{0}^{\pi} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx + 2\pi \int_{0}^{\pi} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx - I I = − π 2 ∫ 0 π sin 4 x + cos 4 x sin x cos x d x + 2 π ∫ 0 π sin 4 x + cos 4 x x sin x cos x d x − I 2 I = − π 2 ∫ 0 π sin x cos x sin 4 x + cos 4 x d x ⏟ = 0 (odd symmetry about π / 2 ) + 2 π ∫ 0 π x sin x cos x sin 4 x + cos 4 x d x 2I = -\pi^2 \underbrace{\int_{0}^{\pi} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx}_{= 0 \text{ (odd symmetry about } \pi/2)} + 2\pi \int_{0}^{\pi} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx 2 I = − π 2 = 0 (odd symmetry about π /2 ) ∫ 0 π sin 4 x + cos 4 x sin x cos x d x + 2 π ∫ 0 π sin 4 x + cos 4 x x sin x cos x d x I = π ∫ 0 π x sin x cos x sin 4 x + cos 4 x d x I = \pi \int_{0}^{\pi} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx I = π ∫ 0 π sin 4 x + cos 4 x x sin x cos x d x Solve the Reduced Integral
Let J = ∫ 0 π x sin x cos x sin 4 x + cos 4 x d x J = \int_{0}^{\pi} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx J = ∫ 0 π s i n 4 x + c o s 4 x x s i n x c o s x d x x → π − x x \to \pi-x x → π − x
J = ∫ 0 π ( π − x ) sin x ( − cos x ) sin 4 x + cos 4 x d x = − π ∫ 0 π sin x cos x sin 4 x + cos 4 x d x + J J = \int_{0}^{\pi} \frac{(\pi-x) \sin x (-\cos x)}{\sin^4 x + \cos^4 x} \, dx = -\pi \int_{0}^{\pi} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx + J J = ∫ 0 π sin 4 x + cos 4 x ( π − x ) sin x ( − cos x ) d x = − π ∫ 0 π sin 4 x + cos 4 x sin x cos x d x + J This confirms J J J π / 2 \pi/2 π /2 x → π / 2 − x x \to \pi/2 - x x → π /2 − x [ 0 , π / 2 ] [0, \pi/2] [ 0 , π /2 ]
J = − π 2 ∫ 0 π / 2 sin x cos x sin 4 x + cos 4 x d x J = -\frac{\pi}{2} \int_{0}^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx J = − 2 π ∫ 0 π /2 sin 4 x + cos 4 x sin x cos x d x Evaluate the Numerical Integral
Let u = tan 2 x u = \tan^2 x u = tan 2 x d u = 2 tan x sec 2 x d x du = 2\tan x \sec^2 x \, dx d u = 2 tan x sec 2 x d x cos 4 x \cos^4 x cos 4 x
∫ 0 π / 2 tan x sec 2 x tan 4 x + 1 d x = 1 2 ∫ 0 ∞ d u u 2 + 1 = 1 2 [ tan − 1 u ] 0 ∞ = 1 2 ⋅ π 2 = π 4 \int_{0}^{\pi/2} \frac{\tan x \sec^2 x}{\tan^4 x + 1} \, dx = \frac{1}{2} \int_{0}^{\infty} \frac{du}{u^2 + 1} = \frac{1}{2} \left[ \tan^{-1} u \right]_0^\infty = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} ∫ 0 π /2 tan 4 x + 1 tan x sec 2 x d x = 2 1 ∫ 0 ∞ u 2 + 1 d u = 2 1 [ tan − 1 u ] 0 ∞ = 2 1 ⋅ 2 π = 4 π So, J = − π 2 ⋅ π 4 = − π 2 8 J = -\frac{\pi}{2} \cdot \frac{\pi}{4} = -\frac{\pi^2}{8} J = − 2 π ⋅ 4 π = − 8 π 2
Final Substitution
I = π J = π ( − π 2 8 ) = − π 3 8 I = \pi J = \pi \left( -\frac{\pi^2}{8} \right) = -\frac{\pi^3}{8} I = π J = π ( − 8 π 2 ) = − 8 π 3 Substitute I I I
∣ 120 π 3 ⋅ ( − π 3 8 ) ∣ = ∣ − 15 ∣ = 15 \left| \frac{120}{\pi^3} \cdot \left( -\frac{\pi^3}{8} \right) \right| = \left| -15 \right| = 15 π 3 120 ⋅ ( − 8 π 3 ) = ∣ − 15 ∣ = 15 The final answer is 15.