integrate 1/sin^4x+sin^2xcos^2x+cos^4x - Maths - Integrals

Question

integrate 1/sin^4x+sin^2xcos^2x+cos^4x

Manbar Singh · Accepted Answer

I =∫ dxsin4x + cos4x + sin2x cos2x⇒I = ∫ dxsin2x
+ cos2x2 - 2 sin2x cos2x + sin2x cos2x⇒I = ∫ dx1 -
sin2x cos2x⇒I = ∫ dx1 - sin2x 1 - sin2x ⇒I =
∫ dx1 - sin2x + sin4x⇒I = ∫ dxcos2x + sin4x

⇒I = ∫ 1cos4x dxcos2xcos4x + sin4xcos4x dividing
numerator and denominator by cos4x⇒I = ∫ sec4xsec2x +
tan4x dx⇒I = ∫sec2x sec2xsec2x + tan4x dx⇒I =
∫ 1 + tan2x sec2x1 + tan2x + tan4x dx

Put tan x = t⇒ sec2x dx= dtTherefore, I = ∫ 1 + t21 +
t2 + t4 dt⇒ I = ∫ t2 + 1 t4 + t2 + 1 dt⇒ I =
∫ t2 1 + 1t2t2 t2 + 1 + 1t2 dt⇒ I = ∫ 1 + 1t2 t2
+ 1t2 + 1 dt

⇒I = ∫ 1 + 1t2t - 1t2 + 2 + 1 dt⇒I = ∫ 1 +
1t2t - 1t2 + 32 dtPut t - 1t = u1 + 1t2 dt = duTherefore, I
=∫ duu2 + 32⇒I = 13tan-1u3 + C⇒I = 13tan-1 t -
1t3 + C

⇒I = 13tan-1t2 - 13t + C⇒I = 13tan-1 tan2x - 13tan x +
C