Question

integration 0 to pi by 2 root over tanx dx​

Answer 1

Step-by-step explanation:

Let I = ∫π/20 tan x−−−−−√ dx ...(1)

⇒I = ∫π/20 tan (π/2−x)−−−−−−−−−−−√ dx

⇒I = ∫π/20cot x−−−−√ dx ...(2)

Adding (1) and (2), we get

2I = ∫π/20 [tan x−−−−−√ + cot x−−−−√] dx

⇒2I = ∫π/20(sin x + cos xsin x . cos x√) dx

⇒2I = 2√∫π/20[sin x + cos x2 sin x . cos x√] dx

⇒2√I = ∫π/20[sin x + cos x1 − (sin x − cos x)2√] dx

Put sin x − cos x = t

⇒(cos x + sin x ) dx = dt

as x→0, t→−1

as x→π2, t→1

⇒2√I = ∫1−1 dt1−t2√

⇒2√I =[sin−1t]+1−1

⇒2√I =sin−1(1) − sin−1(−1)

⇒2√I = π2 + π2

⇒2√I =π

⇒I = π2√