Another Explanation (5):
∫0π/4 (tan3x + tan x) dx = ?
সমাধান:
ধরি, I = ∫0π/4 (tan3x + tan x) dx
= ∫0π/4 tan x (tan2x + 1) dx
= ∫0π/4 tan x sec2x dx
এখন, ধরি tan x = t
সুতরাং, sec2x dx = dt
যখন x = 0, t = tan 0 = 0
যখন x = π/4, t = tan (π/4) = 1
অতএব, I = ∫01 t dt
= [t2/2]01
= (12/2) - (02/2)
= 1/2 - 0
= 1/2
সুতরাং, ∫0π/4 (tan3x + tan x) dx = 1/2 😊
