integrate: limits 0 to pi/2 x sinx cosx /(sin4x +cos4x) - Maths - Integrals

Question

integrate: limits 0 to pi/2 x sinx cosx /(sin4x
+cos4x)

Manbar Singh · Accepted Answer

Let I = ∫0π/2 x sin x cos xsin4x + cos4x dx     
1⇒I = ∫0π/2π/2-x
 sinπ/2-x   cosπ/2-xsin4π/2-x 
 cos4π/2-x dx⇒I = ∫0π/2π/2-x sin x cos xcos4x + sin4x dx   
2adding 1 and 2, we get2I = π2∫0π/2  sin x cos xsin4x + cos4x dx⇒I = π4∫0π/2  sin x cos xsin2x2 + 1-sin2x2 dxput sin2x = t⇒2 sin x 
 cos x dx = dt⇒ sin x 
 cos x dx = dt2as x→0, t→0as x→π/2, t→1Now, I = π8∫01dtt2 + 1-t2⇒I = π8∫01dt2t2-2t+1⇒I = π16∫01dtt2-t+12⇒I = π16∫01dtt2-t+14+14⇒I = π16∫01dtt-122+122⇒I = π16×2tan-1t-121/201⇒I = π8tan-12t-101⇒I = π8tan-11 - tan-1-1⇒I = π8tan-11+tan-11⇒I = π8π4+π4⇒I = π8×π2⇒I = π216