2Weierstrass substitution in integration
put t = tan (x/2)---------------------A
sin x = 2 t / (1 + t²)----------------B
cos x = (1 - t²) / (1 + t²)---------C
dx=2 dt / (1+ t²)---------------------D
When t= tan x/2 , sin x= ? cos x=? dx=? All these doubts are cleared below
dt/dx (tan x/2) = sec²( x/2) • (1/2)= (1/2) • (√t² + 1)² = (t² + 1)/2
{ dy/dx tan x = sec² x }
∴ dx= dt/(t² + 1)/2 = 2 dt/(t² + 1)---------------------------D
As tan x/2 = t, in the right angle triangle, opp side= t, adj side = 1 and hyp =(√t²+1)
∴sin x/2 = t/√t²+1 and cos x/2 = 1/√t²+1
We now have to convert sin x/2 to sin x and cos x/2 to cos x
sin x = sin ( 2 • x/2 ) = 2 sin x/2 • cos x/2 { sin 2x = 2 sin x cos x )
= 2 ( t/√t² + 1 ) (1/√t² + 1 ) = 2 t / (t² + 1)
∴ sin x = 2 t/(t² + 1)------------------------------------------------B
cos x = cos ( 2 • x/2 )= 2 cos² x/2 - 1 { cos 2x = 2 cos²x — 1 }
= 2 { (1/√t² + 1) }² —1
= 2/ ( t² + 1 ) — 1= 2 —( t² + 1 )/(t² + 1 )
= (1 - t²)/(1 + t²)
∴ cos x= (1— t²)/(1 + t²)-----------------------------------------C
