Methods of Integration

Transcription

1 Methods of Integration

2 Essential Formulas k d = k +C sind = cos +C n d = n+ n + +C cosd = sin +C e d = e +C tand = ln sec +C d = ln +C cotd = ln sin +C + d = tan +C lnd = ln +C secd = ln sec + tan +C cscd = ln csc cot +C

3 Substitution f (g()) g ()d = f (u)du Eamples: u = g() du = g ()d + 4 d = / u du u = + 4 = ln u +C du = d = ln( + 4) +C + d = + d = (u ) u du u = + du = d = (u / u / )du When looking for an appropriate substitution, look for compositions of functions. Try letting the inside function be the u. Remember that when substituting, everything must be accounted for, all s must be converted to u s and that there must be a du. = 5 u5/ u/ +C = 5 ( + ) 5/ ( + ) / +C

4 Integration by Parts udv = uv vdu Eample: e d Integration by parts is useful when the integrand is a product of two different kinds of pieces. For instance, an eponential term times a trigonometric term, or a logarithmic term times an algebraic term. e d = e e d = e e e d = e e e = e e + 6e 6e +C e d u = du = d u = du = d u = du = d dv = e d v = e dv = e d v = e dv = e d v = e Note that it may be necessary to do the procedure more than once. If after a few iterations you end up back at the starting integral, you may be able to solve the integral by gathering the occurrences of that integral on one side of the equation. Rule of Thumb for choosing u (epressions at the top of the list tend to make better choices of u): Inverse trigonometric Logarithmic Algebraic Trigonometric Eponential epression

5 Trigonometric Integrals Eamples: sin 5 cos d = sin sin 4 cos d = sin( cos ) cos d sin m cos n d = sin(cos cos 4 + cos 6 )d = ( u + u 4 u 6 )du u = cos du = sind = u + u5 5 u7 7 +C = cos + cos5 5 cos7 7 +C If n is odd, use the identity sin + cos =, to convert all but one cosine terms to sine. Then substitute u = sin. Similarly, if m is odd, convert the sine terms to cosine, leaving one sine term, and substitute u = cos. If both powers are even, then you must use a combination of double angle identities to simplify the integrand. Begin with: sin()cos() = sin() sin 4 cos 4 d = sin() 4 d = 6 sin 4 ()d = 6 [ cos()] d = 64 = 64 cos() + cos () d sin() + ( + cos(4))d Then use: sin () = [ cos()] cos () = [ + cos()] = 64 = 64 sin() + + sin(4) +C 4 sin() + 8 sin(4) +C

6 Eample: tan m sec n d tan sec d If n is even, use the identity sec = + tan to convert all but two of the secants into tangents. Then substitute u = tan. If m is odd, convert all but one of the tangents into secant, and substitute u = sec. Similar strategies work for combinations of powers of cotangent and cosecant. tan sec d = tan(sec )sec d = (sec sec)sectand = (u u)du = u4 4 u +C = sec4 4 sec +C u = sec du = sectand multiple angles Use the identities (derived from addition formulas for sine and cosine): sin(m) sin(n) d cos(m) cos(n) d sin(m) cos(n) d sinasinb = [sin(a B) cos(a + B)] cosacosb = [cos(a B) + cos(a + B)] sinacosb = [sin(a B) + sin(a + B)]

7 Trigonometric Substitution Eamples: θ θ θ a a a a a a d (/)sec = θ dθ secθ = d = secθ dθ = ln secθ + tanθ +C = 9 ln C = d ( + ) + du u + 9 sec θ = secθ dθ = secθ dθ 9 4 = tanθ = secθ d = sec θ dθ u = + du = d u 9 u Trigonometric substitution is useful when the integrand has a term of the form + a, a, or a. This term is often (but not always) inside of a square root or in the denominator of a fraction. There are essentially three cases, all involving replacing algebraic epressions with trigonometric epressions. Note that it may be necessary to complete the square. = ln secθ + tanθ +C u = ln u +C ( + ) = ln C u = tanθ u + 9 = secθ du = sec θ dθ

8 Partial Fractions If the degree of the numerator is higher than the degree of the denominator, before beginning the partial fractions procedure, you must perform polynomial long division. Eample: d The first step in determining the partial fractions decomposition is to factor the denominator. While in practice this may be very difficult, in theory it is possible to factor the denominator into the following types of forms: + a = A B +C + + = A( + ) + (B +C) + + a + b 4 ( + a) n ( + a + b) n The corresponding terms in the partial fraction decomposition: 5 + = A + A + B +C 5 = A + B = = A = B = 0 = C A + a A + B + a + b A + a + A ( + a) + + A n ( + a) n d = + d + du = ln + u = ln + ln u +C u = + du = d 4 A + B + a + b + A + B ( + a + b) + + A n + B n ( + a + b) n = ln + ln( + ) +C Use linear algebra to find the values of the coefficients in the numerators of these fractions.