Let x = a tan so that θ θ a x 49 Then x2 + a2 = a2 tan2 + a2 = a2 (1 + tan2 = a2 sec2 θ θ) θ dx = d (a tan = a sec2 d θ) θ θ = 1 Hence   2 sec a 1 d = 2 dx a 1 = x a tan  c a Example 13 dx Prove x x a 2 2 sin   c a Proof Let x = a sin so that dx = a cos d θ θ θ sin   a x a a and = 2 2 2 2 2 cos  a = cos  cos   a a 2 2 dx cos d a Hence x a 2 2  cos  a = d   = c x sin = a  c x = a sin θ Example 14 2 4 Find =  dx
Let x = a tan θ so that θ x a 49 Then x2 + a2 = a2 tan2 θ + a2 = a2 (1 + tan2 θ) = a2 sec2 θ dx = d (a tan θ) = a sec2 θ d θ Hence 2 dx  = = a sec 2 d 1 1  a = 1 a tan x a  c Example 13 Prove dx x 2 a 2  sin x a  c Proof Let x = a sin θ so that dx = a cos θ d θ and a 2 x 2 = a 2 a 2 sin 2   = a cos  a 2 cos 2   a cos  Hence dx x 2 a 2  a cos d a cos  = = d   c = sin x a  c x = a sin θ Example 14 Find = dx 2  4
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