Question

Integration sin2x /sin^4x + cos^4x dx​

Answer 1

Answer:

Step-by-step explanation:

∫ [ ( sin 2x ) / ( sin⁴ x + cos⁴ x ) ] dx

= ∫ [ ( 2 sin x cos x ) / ( sin⁴ x + cos⁴ x ) ] dx

Divide numerator and denominator both by cos⁴ x :

= ∫ [ ( 2 sin x / cos³ x ) / ( tan⁴ x + 1 ) ] dx

= ∫ [ 2 ( sin x / cos x )( 1 / cos² x ) / ( 1 + tan⁴ x ) ] dx

= ∫ { 1 / [ 1 + ( tan² x )² ] } • 2 tan x sec² x dx

= ∫ [ 1 / ( 1 + u² ) ] du, ………………

u = tan² x, du = 2 tan x sec² x dx

= [ tanֿ¹ ( u ) ] + C

= [ tanֿ¹ ( tan² x ) ] + C