# integrate root tanx

Rahul Kumar
131 Points
12 years ago

tan x = sin x / cos x
Put y = cos x
Now dy = d(cos x) = -sin x dx
Hence, int (tan x dx) = int (sin x dx / cos x) = int (-dy / y)
= - ln y
= -ln (cos x)
= ln (sec x)

aniket deroy
18 Points
12 years ago

INTEGRATION NOT POSSIBLE

Qwerty
11 Points
6 years ago
∫√(tan x) dxLet tan x = t2⇒ sec2 x dx = 2t dt⇒ dx = [2t / (1 + t4)]dt⇒ Integral ∫ 2t2 / (1 + t4) dt⇒ ∫[(t2 + 1) + (t2 - 1)] / (1 + t4) dt⇒ ∫(t2 + 1) / (1 + t4) dt + ∫(t2 - 1) / (1 + t4) dt⇒ ∫(1 + 1/t2 ) / (t2 + 1/t2 ) dt + ∫(1 - 1/t2 ) / (t2 + 1/t2 ) dt⇒ ∫(1 + 1/t2 )dt / [(t - 1/t)2 + 2] + ∫(1 - 1/t2)dt / [(t + 1/t)2 -2]Let t - 1/t = u for the first integral ⇒ (1 + 1/t2 )dt = duand t + 1/t = v for the 2nd integral ⇒ (1 - 1/t2 )dt = dvIntegral= ∫du/(u2 + 2) + ∫dv/(v2 - 2)= (1/√2) tan-1 (u/√2) + (1/2√2) log(v -√2)/(v + √2)l + c= (1/√2) tan-1 [(t2 - 1)/t√2] + (1/2√2) log (t2 + 1 - t√2) / t2 + 1 + t√2) + c= (1/√2) tan-1 [(tanx - 1)/(√2tan x)] + (1/2√2) log [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c