JUST THE MATHS SLIDES NUMBER 12.8 INTEGRATION 8 (The tangent - PDF document

“JUST THE MATHS” SLIDES NUMBER 12.8 INTEGRATION 8 (The tangent substitutions) by A.J.Hobson 12.8.1 The substitution t = tan x 12.8.2 The substitution t = tan( x/ 2)

UNIT 12.8 - INTEGRATION 8 THE TANGENT SUBSTITUTIONS 12.8.1 THE SUBSTITUTION t = tan x This substitution is used for integrals of the form 1 � a + b sin 2 x + c cos 2 x d x, where a , b and c are constants. In most exercises, at least one of these three constants will be zero. A simple right-angled triangle will show that, if t = tan x , then t 1 sin x ≡ 1 + t 2 and cos x ≡ 1 + t 2 . √ √ √ ✟ ✟✟✟✟✟✟✟✟✟✟✟ 1 + t 2 t x 1 Furthermore, d x ≡ sec 2 x ≡ 1 + t 2 so that d x d t 1 1 + t 2 . d t ≡ 1

EXAMPLES 1. Determine the indefinite integral 1 � 4 − 3sin 2 x d x. Solution 1 � 4 − 3sin 2 x d x 1 1 � = . 1 + t 2 d t 4 − 3 t 2 1+ t 2 1 � = 4 + t 2 d t = 1 2tan − 1 t 2 + C = 1  tan x   2tan − 1  + C.   2 2. Determine the indefinite integral 1 � sin 2 x + 9cos 2 x d x. 2

Solution 1 � sin 2 x + 9cos 2 x d x 1 1 � = . 1 + t 2 d t t 2 9 1+ t 2 + 1+ t 2 1 � = t 2 + 9 d t = 1 3tan − 1 t 3 + C = 1  tan x   3tan − 1  + C.   3 12.8.2 THE SUBSTITUTION t = tan( x/ 2) This substitution is used for integrals of the form 1 � a + b sin x + c cos x d x, where a , b and c are constants. In most exercises, one or more of these constants will be zero In order to make the substitution, we make the following observations: 3

(i) sin x ≡ 2 sin( x/ 2) . cos( x/ 2) ≡ 2 tan( x/ 2) . cos 2 ( x/ 2) ≡ 2 tan( x/ 2) 2 tan( x/ 2) 1 + tan 2 ( x/ 2) . sec 2 ( x/ 2) ≡ Hence, 2 t sin x ≡ 1 + t 2 . (ii) cos x ≡ cos 2 ( x/ 2) − sin 2 ( x/ 2) ≡ cos 2 ( x/ 2) 1 − tan 2 ( x/ 2) � � ≡ 1 − tan 2 ( x/ 2) ≡ 1 − tan 2 ( x/ 2) 1 + tan 2 ( x/ 2) . sec 2 ( x/ 2) Hence, cos x ≡ 1 − t 2 1 + t 2 . (iii) d x ≡ 1 d t 2sec 2 ( x/ 2) ≡ 1 ≡ 1 1 + tan 2 ( x/ 2) 2[1 + t 2 ] . � � 2 4

Hence, d x 2 1 + t 2 . d t ≡ EXAMPLES 1. Determine the indefinite integral 1 � 1 + sin x d x. Solution 1 � 1 + sin x d x 1 2 � = . 1 + t 2 d t 2 t 1 + 1+ t 2 2 � = 1 + t 2 + 2 t d t 2 � = (1 + t ) 2 d t = 2 2 1 + t + C = − 1 + tan( x/ 2) + C. − 2. Determine the indefinite integral 1 � 4 cos x − 3 sin x d x. 5

Solution 1 � 4 cos x − 3 sin x d x 1 2 � = . 1 + t 2 d t 4 1 − t 2 6 t 1+ t 2 − 1+ t 2 2 � = 4 − 4 t 2 − 6 t d t 1 � = 2 t 2 + 3 t − 2 d t − 1 � (2 t − 1)( t + 2) d t − � 1 1 2    d t = t + 2 −    5 2 t − 1 = 1 5[ln( t + 2) − ln(2 t − 1)] + C = 1  tan( x/ 2) + 2   5 ln  + C.     2 tan( x/ 2) − 1 6