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integration of rational function of sine and cosine∗ pahio† 2013-03-21 22:52:49 The integration task Z R(sin x, cos x) dx, (1) where the integrand is a rational function of sin x and cos x, changes via the Weierstrass substitution x tan = t (2) 2 to a form having an integrand that is a rational function of t. Namely, since x = 2 arctan t, we have dx = 2 · 1 dt, 1+t2 2t , 1+t2 cos x = (3) and we can substitute sin x = 1−t2 , 1+t2 (4) getting Z Z R(sin x, cos x) dx = 2 R 2t 1−t2 , 2 1+t 1+t2 dt . 1+t2 Proof of the formulae (4): Using the double angle formulas of sine and cosine and then dividing the numerators and the denominators by cos2 x2 we obtain sin x = 2 sin x2 cos x2 2 tan x2 2t = 2 x 2 x = 1 + t2 , x 2 1 + tan sin 2 + cos 2 2 cos x = cos2 x2 − sin2 sin2 x2 + cos2 x 2 x 2 = 1 − tan2 1 + tan2 ∗ hIntegrationOfRationalFunctionOfSineAndCosinei x 2 x 2 = 1 − t2 . 1 + t2 created: h2013-03-21i by: hpahioi version: h39380i Privacy setting: h1i hTopici h26A36i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Z Example. The above formulae give from Z dx = sin x Z 1+t2 1 dt = ·2· 2t 1+t2 Z dx the result sin x dt x = ln |t| + C = ln tan + C t 2 (which can also be expressed in the form − ln | csc x + cot x| + C; see the goniometric formulas). Note 1. The substitution (2) is sometimes called the “universal trigonometric substitution”. In practice, it often gives rational functions that are too complicated. In many cases, it is more profitable to use other substitutions: R • In the case R(sin x) cos x dx the substitution sin x = t is simpler. R • Similarly, in the case R(cos x) sin x dx the substitution cos x = t is simpler. • If the integrand depends only on tan x, the substitution tan x = t is simpler. • If the integrand is of the form R(sin2 x, cos2 x), one can use the substitution tan x = t; then 1 1 t2 dt 2 2 , sin x = 1−cos x = , dx = . cos2 x = = 2 2 2 1 + t 1 + t 1 + t2 1 + tan x Z dx Example. The integration of dx is of the last case: cos4 x Z Z Z Z dx 1 dt t3 1 2 2 2 dx = dx = (1+t ) · = (1+t ) dt = +t+C = tan3 x+tan x+C. 4 2 2 2 cos x (cos x) 1+t 3 3 Z Z dx Example. The integral I = dx = sec3 x dx is a peculiar case in cos3 x which one does not have to use the substitutions mentioned above, as integration by parts is a simpler method for evaluating this integral. Thus, u = sec x ⇒ du = sec x tan x dx; dv = sec2 x dx ⇒ v = tan x. Therefore, Z I = sec3 x dx Z = sec x tan x − Z sec x tan2 x dx sec x (sec2 x − 1) dx Z = sec x tan x − I + sec x dx, = sec x tan x − 2 and consequently Z dx 1 dx = sec x tan x + ln | sec x + tan x| + C. 3 cos x 2 Note 2. There is also the “universal hyperbolic substitution” for integrating rational functions of hyperbolic sine and cosine: tanh x = t, 2 dx = 2dt , 1−t2 sinh x = 2t , 1−t2 cosh x = 1+t2 1−t2 References [1] Л. Д. Кдрячев: Математичецкии анализ. Издательство “ВүсшаяШкола”. Москва (1970). 3