Integral of
(tan x)1/2 + (cot x)1/2
.Q. Intergrate : √ (tanx) + √(cotx)
ANS : I = ∫ { √ (tanx) + √(cotx) } dx
Apply : tanx = sinx / cosx and cotx = cosx / sinx
=> ∫ { √ (sinx/ cosx) + √(cosx / sinx) } dx
=> ∫ { √ (sin2x) + √(cos2x) } dx / √(cosx *sinx)
=> ∫ ( sinx + cosx ) dx / √( sinx cosx )
=> √2 * ∫ ( sinx + cosx ) dx / √( 2 * sinx cosx )
=> √2 * ∫ ( sinx + cosx ) dx / √( 1-1 + 2 * sinx cosx )
=> √2 * ∫ ( sinx + cosx ) dx / √( 1- { 1 - 2 * sinx cosx} )
=> √2 * ∫ ( sinx + cosx ) dx / √( 1- { sin2x + cos2x - 2 * sinx cosx} )
=> √2 * ∫ ( sinx + cosx ) dx / √( 1- { sinx - cosx}2 )
Let t = sin x - cos x => dt = (cosx + sinx ) dx => dt = (sinx + cosx) dx
∴ √2 * ∫ dt / √( 1- { t }2 )
Apply : ∫ [1 / √( 1- x2 ) ] dx = sin-1x + C
=> √2 * sin-1t + C
∴ I = √2 * sin-1(sin x - cos x )+ C
Therefore : ∫ { √ (tanx) + √(cotx) } dx = √2 * sin-1(sin x - cos x )+ C
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