root tanx integration ( limits 0 to pi/2) Share with your friends Share 7 Mayur Pisode answered this Let I = ∫0π/2 tan x dx ...1⇒I = ∫0π/2 tan π/2-x dx ⇒I = ∫0π/2cot x dx ...2Adding 1 and 2, we get2I = ∫0π/2 tan x + cot x dx⇒2I = ∫0π/2sin x + cos xsin x . cos x dx⇒2I = 2∫0π/2sin x + cos x2 sin x . cos x dx⇒2I = ∫0π/2sin x + cos x1 - sin x - cos x2 dxPut sin x - cos x = t⇒cos x + sin x dx = dtas x→0, t→-1as x→π2, t→1⇒2I = ∫-11 dt1-t2⇒2I =sin-1t-1+1⇒2I =sin-11 - sin-1-1⇒2I = π2 + π2⇒2I =π⇒I = π2 13 View Full Answer
