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Calculus study notes from Examville - Integration
Examville is a global education community where users like you can connect and interact with other students and teachers from around the world. Share, seek, download and discuss everything inside and outside the classroom. All you need is an email address and a password to get started. JOIN US FOR FREE: http://www.examville.com © Examville.com, LLC June 2009 INTEGRATION Integration or the process of finding the antiderivative is one of the most important operations in calculus. It is the opposite process of differentiation. ANTIDERIVATIVE OR INDEFINITE INTEGRAL is the derivative of the function F x , that is F. x = f x for all x in the ` a ` a ` a ` a If f x ` a ` a domain of f, then F x is called the antiderivative or indefinite integral of f x . We write Z f x dx = F x + C ` a ` a F x = x 2 @ 2, ` a The antiderivative of a given function is not unique. For example, G x = x 2 + 5, and H x = x2 f x = 2x ` a ` a ` a are all antiderivatives of as dff c d b c d b c ff ff F. x = G. x = H. x = f x or ff x 2 @ 2 = ff x 2 + 5 = ff x 2 = 2x for all x in the ff ff ff ff ff ff ` a ` a ` a ` a b dx dx dx domain of f . We note that, all the antiderivatives of the function f differ only by a constant (the derivative of the constant value is always zero). Geometrically, this means that the graphs ` a ` a ` a of F x , G x and H x are identical except for their vertical position. All their derivatives will be same for any x = x 0 . All indefinite integrals of f x = 2x are then ` a expressed by the general form of the antiderivative which is F x = x 2 + C where C is the ` a constant of integration or an arbitrary constant. The notation Z is used for integration. The symbol Z f x dx is used to denote the ` a indefinite integral of the function f(x). The function f(x) is called the integrand. So, we can write Z 2x dx = x 2 + C . The indefinite integral of a function is sometimes called the general antiderivative of the function. INTEGRATION BY INSPECTION We shall solve the first few questions by method of inspection, that is, comparing the integrand with some known derivatives. f x = sin x ` a Example : Find the indefinite integral of the function Solution : F x = @ cos x is F. x = sin x ` a ` a We know that the derivative of the function so we write Z sin x dx = @ cos x + C f x = x2 ` a Example : Find the indefinite integral of the function Solution : f 3g dff c ff ff ff 3 ff xff dff ff ff ff ff f 1f ff x = 3x , or = A 3x 2 = x 2 b 2 We know that dx dx 3 3 3 xff ff f ff As the derivative of F x = is F. x = x 2 ` a ` a 3 3 xff ff f Z x 2 dx = ff+ C 3 STANDARD INTEGRATION FORMULAS: COMPARISON WITH DIFFERENTIATIONS In the above cases of integration by inspection, the antiderivatives were found by comparing the integrand with some known derivatives. But it is not always possible to do so, as in most cases the integrand does not match with our known derivatives. So, we need some integration formulas. As integration is the reverse process of differentiation, we can make the first few integration formulas directly from the corresponding derivative formulas. These are given below : For Polynomial Functions: Differentiation Formula Corresponding Integration Formula n+1 dff ff` ff ff xffff xn = nA xn@1 ffff ffff Z x n dx = fffff C + a dx n+1 dff ff ff ff` a x = 1 Z dx = x + C dx For Trigonometric Functions: Differentiation Formula Corresponding Integration Formula dff ff ff ff` sin x = cos x Z cos x dx = sin x + C a dx dff ff ff ff` cos x = @ sin x Z sin x dx = @ cos x + C a dx dff ff ff ff` tan x = sec 2 x Z sec 2 x dx = tan x + C a dx dff ff ff ff` cot x = @csc 2 x Z csc 2 x dx = @ cot x + C a dx dff ff ff ff` sec x = sec x tan x Z sec x tan x dx = sec x + C a dx dff ff ff ff` csc x = @ csc x cot x Z csc x cot x dx = @ csc x + C a dx For Exponential & Logarithmic Functions: Differentiation Formula Corresponding Integration Formula dff ff ff ff` x a e = ex Z e x dx = e x + C dx x dff ff` x a ff ff afff a = a x ln a fff fff Z a x dx = fff+ C a>0, a ≠ 1 dx ln a dff ff ff` ff a 1f 1f ln x = fff f f Z fdx = ln |x| + C dx x x INTEGRATION OF COMBINATION OF FUNCTIONS The following rules of integration for addition, subtraction and scaler multiples of functions can be used. Note that, unlike differentiation, product and quotients of functions are not covered in these (which are to be solved using substitution or integration by parts or by some other methods). Z c f x dx = c Z f x dx ` a ` a Z f x F g x dx = Z f x dx F Z g x dx B ` a ` aC ` a ` a Example : Evaluate Z x 3 dx Solution : 3+1 xffff 1f ffff fff f f Z x 3 dx = ffff+ C = fx 4 + C 3+1 4 1f fff fff fff f Example : Evaluate Z pwf w dx w w ww w x Solution : 1 1f @ f+ 1 f f 1f ff ff xfffff w ww w w ww fff Z fwf = Z x @ 2 dx = ffffff C = 2 p x + C w dx w ff fff f w ww w ff fff ff2fff f + px 1f @ +1 fff 2 SUBSTITUTION METHOD: CHANGE OF VARIABLES Till now we have discussed integrals of those functions which are readily obtained from derivatives of some known functions. They could be solved using the standard formulas. But some integrals can not be evaluated directly as no standard formula matches the form wwww wwww wwww www www www www www ` a2 fx ff lnffff ffff ff fffff of the given function. Some examples are Z x q x 2 + 5 dx , Z dx , x Z tan3 x sec 2 x dx . One way of solving them is to substitute some new variables in the integral and thus transforming it to standard form with the new variable. After choosing the suitable variable, we have to rewrite the integral in terms of the new variable, so that one or more of the standard formulas can be used. After the integration process is over, the result is to be written in terms of the original variable. c4 Example : Evaluate Z 3x x 2 + 1 dx b Solution : Let t = x 2 + 1 dtf ff ff f ff [ = 2x dx 1f [ f x dx = fdt 2 4 3f f Z 3x x 2 + 1 dx = Z ft 4 dt b c 2 5 3f tff f f f f 3f = ff ff A f C = fft 5 + C + 2 5 10 3fb c5 ff ff = ff x 2 + 1 + C 10 Example : Evaluate Z sin 3x dx Solution : Let 3x = t 1f ff dx = dt 3 1f f Z sin 3x dx = Z fsin t dt 3 1f ff 1f ff = @ cos t + C = @ cos 3x + C 3 3 Example : Evaluate Z tan2 2x @ 3 dx ` a Solution : Let 2x @ 3 = t 1f ff dx = dt 2 1f f Z tan2 2x @ 3 dx = Z ftan2 t dt ` a 2 H I 1f b ff 1f ff = Z sec 2 t @ 1 dt = J Z sec 2 t dt @Z dt K c 2 2 1f ff = tan t @ t + C @ A 2 1f ` ff = tan 2x @ 3 @ 2x @ 3 + C B a ` aC 2 OTHER INTEGRATION FORMULAS Apart from the standard formulas given above, the following formulas are also used in integration (these are obtained by using the substitution method) : Z tan x dx = @ ln |cos x| + C Z cot x dx = ln |sin x| + C Z sec x dx = ln |sec x + tan x| + C Z csc x dx = @ ln|csc x + cot x| + C dx ffffff 1f fffff Z ffffff = farctan f + C fffff ff xf f f x +a 2 2 a a L M dx @ aM 1ff L xffff ffffff ffffff fffff Z ffffff = ff lnL fffff + C ff ff ffff fffff M 2 2 2a L x + a M L M x @a + xM L M dx fffff ffffff ffffff ffffff ff L affff 1ff L fffff ff ff ffff ffff Z 2 = lnL M+ M C a @ x2 2a La @ xM wwww wwww wwww wwww wwww wwww wwww wwww M dx L ffffffff ffffffff fffffff ffffff Z fwwwwf = lnLx + q x 2 + a 2 M + C L M wwww wwww wwww wwww wwww wwww wwww q x2 + L M a2 wwww M wwww wwww wwww wwww wwww wwww wwwww dx L ffffffff fffffff fffffff fffff Z ffwwwff= lnLx + q x 2 @ a 2 M + C L M wwww wwww wwww wwww wwww wwww w wwwww q x2 @ a2 L M dx xf ffffffff fffffff fffffff fffff Z ffwwwff= arcsin f+ C wwww wwww wwwww wwww wwww wwww wwww w f f qa 2 @ x 2 a wwwww wwwww wwww wwww wwww wwww wwww wwwww wwwww wwwww wwwww wwww wwww wwww wwww wwww wwww wwww wwww wwww wwww wwww wwww wwww M 1f 1f L f f f f Z q x 2 + a 2 dx = fx q x 2 + a 2 + fa 2 lnLx + q x 2 + a 2 M + C L M 2 2 L M wwwww wwwww wwwww wwww wwww wwww wwww wwww wwwww wwwww wwwww wwww wwww wwww wwww wwww wwww M wwww wwww wwww wwww wwww wwww wwwww 1f q 2 1f 2 L L f ff f ff Z q x 2 @ a 2 dx = 2 x x @ a @ a lnLx + q x2 @ a2 M + M C 2 2 L M wwwww wwwww wwww wwww wwww wwww wwww wwwww wwwww wwwww wwwww wwww wwww wwww wwww wwww 1f 1f xf f f f f f f Z q a 2 @ x 2 dx = fx q a 2 @ x 2 + fa 2 arcsin f + C 2 2 a dx ffffffffff 1f ffffff Z fffwwwwf = farcsec f+ C fffffffff fffffffff f f xf f f wwww wwww wwww wwww wwww wwww wwwww x q x2 @ a2 a a INTEGRATION BY PARTS Another useful integration technique for indefinite integrals which do not fit in the basic formulas is integration by parts. We may consider this method when the integrand is a product of two functions. When u and v are differentiable functions of x which is the variable of integration, then d uv = u dv + v du, u dv = d uv @ v du ` a ` a or Integrating both sides we get the following formula for Integration by parts, Z u dv = uv @ Z v du While using this method of integration, the given integral is to be separated into two parts, one part being u and the other part, combined with dx, being dv. For this reason this method is called integration by parts. (a) We have to first choose dv which must be readily integrable to find v. The u function will be the remaining part of the integrand that will be differentiated to find du. (b) The purpose is to find an integral Z v du which is easier to integrate than the original integral Z u dv . Example : Evaluate Z x sec 2 x dx Solution : Let u = x so du = dx and dv = sec 2 x dx then v =Z dv =Z sec 2 x dx = tan x using Z u dv = uv @ Z v du hence Z x sec 2 x dx = x tan x @Z tan x dx = x tan x @ @ ln |cos x| + C b c = x tan x + ln |cos x| + C Example : Evaluate Z ln x 2 + 4 dx b c Solution : 2x Let u = ln x 2 + 4 fffff ffff fffff du = fffffdx b c so x +4 2 and dv = dx then v =Z dv =Z dx = x using Z u dv = uv @ Z v du 2 2x ff fffff fffff ff Z ln x 2 + 4 dx = x ln x 2 + 4 @ Z fffffdx b c b c 2 x +4 8 G fffff fffff ffff = x ln x 2 + 4 @ Z 2 @ fffff dx b c F x +4 2 8 G fffff fffff ffff = x ln x 2 + 4 @ Z 2 @ fffff dx b c F x +4 2 xf f ff = x ln x 2 + 4 @ 2x + 4 arctan + C b c 2 INTEGRATION USING TRIGONOMETRIC IDENTITIES When the integrand involves trigonometric functions which can not be solved by the basic integration formulas, we need to use the different trigonometric identities (together with substitution if needed) to get them into a form in which the basic integration formula scan be applied. The trigonometric identities often used for integration are 1 1f 1 ft sinff ffff fff fff sec t = ffff csc t = fff fff fff f fff fff fff cot t = ffff ffff fff tan t = ffff cos t sin t tan t cos t sin t + cos 2 t = 1 2 tan2 t + 1 = sec 2 t cot 2 t + 1 = csc 2 t @ cos 2α 1fffffffff ffffffff ffffffff sin α = fffffffff 2 2 + cos 2α 1ffffffff ffffffff ffffffff cos 2 α = fffffffff 2 1f sinαcosα = fsin 2αf 2 1f b D cE ff sinαcosβ = sin α + β + sin α @ β c b 2 1f b D cE f cosαsinβ = f sin α + β @ sin α @ β c b 2 1f b D cE f f c b cosαcosβ = cos α + β + cos α @ β 2 1f b D cE f sinαsinβ = f cos α @ β @ cos α + β c b 2 3 Example : Evaluate Z sin x cos 3 x dx Solution : Z sin3 x cos 3 x dx = Z sin3 x cos 2 x A cos xdx = Z sin x 1 @sin x A cos x dx let sin x = t, cos x dx = dt b c B C 3 2 = Z t 3 @ t 5 dt b c 1f 4 1f 6 f ff ff = t @ t +C 4 6 1f 4 f ff 1f 6 ff = sin x @ sin x + C 4 6 Example : Evaluate Z se