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Integration Key Facts Facts which are in the formula book Standard Integrals you need to learn: 1 x dx n 1 x e dx e c n x n1 c x 1 x dx ln x c cos xdx sin x c sin xdx cos x c sec xdx tan x c sec x tan xdx sec x c cos ec xdx cot x c cos ecx cot xdx cos ecx c 2 2 Further generalisations using the above integrals: (ax b) n dx 1 (ax b) n1 c a( n 1) 1 axb e c a 1 1 ax b dx a ln( ax b) c 1 cos(ax b)dx a sin( ax b) c 1 sin( ax b)dx a cos(ax b) c 1 2 sec (ax b)dx a tan( ax b) c 1 sec(ax b) tan( ax b)dx a sec(ax b) c 1 2 cos ec (ax b)dx a cot(ax b) c 1 cos ec(ax b) cot axdx a cos ec(ax b) c e ax b dx Area under a curve, where y = f(x) between x=a and x=b: b ydx a Volume of revolution formed by rotating y about the x axis, where y = f(x) between x=a and x=b: b y 2dx a General patterns you should remember: f ' ( x ) f ( x ) dx n 1 f ( x ) f ' ( x )e dx e c 1 n f ( x) n1 c f ( x) f ' ( x) dx ln f ( x ) c f ( x) f ' ( x ) cos f ( x )dx sin f ( x ) c f ' ( x ) sin f ( x )dx cos f ( x ) c f ' ( x ) sec f ( x )dx tan f ( x ) c 2 Integrating trigonometric powers 1 (1 cos 2 x )dx (double angle formula used) 2 1 2 cos xdx 2 (1 cos 2 x )dx (double angle formula used) 1 n n1 sin x cos xdx n 1 sin x c 1 n n1 cos x sin xdx n 1 cos x c 1 n n sec x tan xdx n sec x c 1 n 2 n1 tan x sec xdx n 1 sec x c sin 2 xdx Integrating using Addition Formula To integrate sin(ax)sin(bx), cos(ax)cos(bx) – cos (A B) = cosAcosB sinAsinB (C3 section) are required. To integrate sin(ax)cos(bx) – sin (A B) = sinAcosB sinAcosB (C3 section) are required. Other trigonometric integrals: tan xdx ln(cos x ) c ln(sec x ) c sec xdx ln(sec x tan x ) c cot xdx ln(sin x ) c cos ecxdx ln(cos ecx cot x ) c Trapezium Rule (C2 section) b f ( x )dx 2 y h a h ba n 0 y n 2( y1 y2 .... yn1 ) Separating variables: dy When f ( x ) g( y ) dx 1 Then dy f ( x )dx g( y ) Integrating parametric equations: When x = f(t), y = (t), the area under the curve: dx y dt dt Volume of revolution formed by rotating y about the x axis, t2 y2 t1 dx dt dt Integration By Parts dv du u dx dx uv v dx c How do you know which is u and dv ? dx Integration by parts is used to integrate a product. One part of the product (u) should be easy to dv differentiate (and will usually simplify when differentiated). The other part should be easy to dx integrate and should not become too much harder when integrated. Examples: Function to be integrated u needs differentiating to find du dx x sin ax x cos ax xn ln x x ex x2ex x2sinx x2cosx lnx x(2x + 3)4 x sec2xtanx x x ln x x x2 parts will need to be applied twice – this would apply to other similar examples. ln x x x Trigonometric Identities which could be required: sin 2 x 12 (1 cos 2 x) cos2 x 12 (1 cos 2 x) sin 2 x cos2 x 1 tan 2 x 1 sec2 x 1 cot 2 x cosec2 x dv needs integrating to find v dx sin ax cos ax xn ex ex sinx cosx 1 (2x + 3)4 sec2xtanx Which method of integration do you use? The methods of integrating an expression are: 1) directly writing it down (either by memory or by looking in the formula book); 2) writing the expression in partial fractions; 3) using the method of integration by parts; 4) using a substitution; 5) using a trigonometric identity (such as sin 2 x 12 (1 cos 2 x) or cos 2 x 12 (1 cos 2 x) ) First look to see if the question tells you (or gives you a hint) about which method of integration you should be using!! First look to see if the integral is easy to work out: Is it one that you should remember (see my sheet) or is it one in the formula book? Also is the integral easier to work out than it looks – e.g. 5x 2 dx x If you are asked to integrate an algebraic fraction of the form ax b or (cx d )(ex f ) ax 2 bx c , you could try writing it in (dx e)2 ( fx g ) partial fractions. (Note: you might be asked in the first part of the question to express it in partial fractions) Examples are: 4 (2 x 1)( x 2) dx x(4 x)dx 2xe dx x cos( x 2)dx (cos x)e dx (2 x 5) ln( x 5 x)dx x2 2 x 2 Use integration by parts to integrate other products, especially when one function is a simple polynomial. Examples are: 3 2sin x 2 To integrate even powers of sin x or cos x , use the formulae sin 2 x 12 (1 cos 2 x) or cos 2 x 12 (1 cos 2 x) . Examples are: cos 4 xdx x dx put u x 2 1 1 sin x put u cos x cos x 2x 1 2 dx put u x 2 x 1 2 ( x x 1) 4 x Recognise that the integral is in the form f '( x) gf ( x) . Examples are: Other quotients (where the denominator cannot be factorised into linear factors) can usually be integrated using a substitution. Examples are: sin 2 xdx (3x 1)e dx x cos xdx 2 x ln xdx 2x 2 3 To integrate odd powers of sin x or cos x , start by using the formula sin 2 x cos 2 x 1 and then use a substitution. Examples are: cos5 xdx sin 3 xdx To integrate sinaxcosbx, use: sin (A +B) = sinAcosB + sinBcosA sin (A -B) = sinAcosB - sinBcosA To integrate sinaxsinbx or cosaxcosbx, use: cos (A -B) = sinAsinB + cosAcosB cos (A +B) = sinAsinB - cosAcosB sin 3 x cos 4 xdx sin 3 x sin 4 xdx Integration involving trigonometric identities Volumes of revolution: rotating about x axis The formula is: Example: 2 cos xdx 12 (1 cos 2 x)dx 12 ( x 12 sin 2 x) c V y 2 dx Example: The diagram shows the graph of 8– 2 2 4 6 2– 2 4 6 8 Find Integration by substitution This column gives the calculations for changing the dx to du: Example: 2 5 x( x 3) dx . Make the substitution u x2 3 We get: xu 5 dx 12 u 5 du This gives: 1 12 sin 3 x cos 4 xdx Solution: We use the sin (4x +3x) = sin4xcos3x + sin3xcos4x sin (4x-3x) = sin4xcos3x – sin3xcos4x y 4 . The region trapped between the lines 2x 1 x = 1, x = 4 and y = 0 is shaded. This region is rotated completely about the x-axis. Find the volume generated. y 8 u x2 3 du 2x Subtracting 1 and 2 dx du 2 xdx sin(4x+3x) – sin(4x-3x) = 2sin3xcos4x So, xdx 12 du u 6 c 121 ( x 2 3)6 6 4 So, sin3xcos4x = ½ sin7x – ½ sinx Example 2: Use the substitution x 1 t 2 x 1 to find dx . 2x 1 sin 3 x cos 4 xdx = – 2 t 2x 1 dt 2 dx dt 2dx dx 12 dt x 1 1 x 1 dx dt Solution: 2 t t Since t 2 x 1 , 2x t 1 x 12 (t 1) So we get 1 (t 1) 2 t dt 1 2 1 4 2 3/ 2 3 t But t 2 x 1 , so 1 4 2 3 This expands to give: (2 x 1) (2 x 1) 1 2 1/ 2 c 6 8 – 2 2 4 Volume = y dx dx 2x 1 1 4 1 dx = 16 1 2x 1 2 0 Use the substitution u x 2 . 4 2e du 2e u u 4 0 = 2e4 2e0 2e4 2 3/ 2 x2 u 0 4 4 Find 4 xe . 2 2 2 Definite integrals using a substitution 2 0 (2 x 1)3/ 2 2(2 x 1)1/ 2 c . 1 6 14 x 4 xe 2t1/ 2 c = 1 cos 7 x 1 cos x c 2 Take the multipliers outside the integral: (t 1) 1 dt 14 (t 1)t 1/ 2 dt 14 t1/ 2 t 1/ 2 dt 4 t This gives: 2 1 1 2 sin 7 x 2 sin xdx u x2 du 2x dx du 2 xdx 4 xdx 2du x=0→u=0 x=2→u=4 4 1 = 16 ln(2 x 1) 2 1 1 1 = 16 ln 7 ln1 8 ln 7 2 2 x 1 4 6 3x 2 c 3 6 2 6 3x 2 c 9 1 1 5 2 x dx 2 ln(5 2 x) 5 4(3x 2) dx 0 0 Example 1: Find 2 xe dx . Common examination questions Example 1: Find x ln xdx . x This is a suitable candidate for integration by parts with u 2 x and du 2 dx dv ex v ex dx u 2x Substitute these into the formula: 2 xe dx 2 xe 2e dx 2 xe x x Example 2: Find x x 2e x c x cos xdx . Here we take u x and This can be found using integration by parts if we dv x. dx du 1 u ln x dx 2 x dv x xv dx 2 dv ex : dx dv cos x : dx du 1 dx dv cos x v sin x dx ux Substitute these into the formula: x cos xdx x sin x sin xdx x sin x ( cos x) c x sin x cos x c Note: Sometimes it is necessary to use the integration by parts formula twice (e.g. with 2 x sin xdx ). 1 1 ln 3 ln 5 2 1 1 5 ln 5 ln 3 ln 2 2 3 Definite integrals (using by parts) Example: a) Find the points where the graph of y (2 x)e x cuts the x and y axes. take u ln x and b) Sketch the graph of y (2 x)e x . c) Find the area of the region between the axes and the graph of y (2 x)e x . a) Graph cuts y-axis when x = 0, i.e. at y = 2 Graph cuts the x-axis when y = 0, i.e. when x = 2.. Substitute these into the formula: x2 1 x2 2 x ln xdx ln x x 2 dx 12 x ln x 12 xdx 2 12 x 2 ln x 14 x 2 c b) The graph looks like: Example 2: Find ln xdx . This can be thought of as 1ln xdx and so can be integrated by parts with u ln x and dv 1 dx du 1 u ln x dx x dv 1 v x dx 1 ln xdx x ln x x x dx x ln x 1dx = x ln x x c y 3 2.5 2 (2 x)e c) Area is x 2 dx . 0 1.5 1 du u 2 x 1 dx dv e x v e x dx 0.5 0.5 1 1.5 So (2 x)e x dx (2 x)(e x ) 1 e x dx = (2 x)(e x ) e x dx = (2 x)(e x ) e x Now that we’ve integrated, we substitute in our limits: 2 (2 x)e 0 x dx (2 x)e x e x 2 2 2 0 0e e 2e e0 0.135 1 1.135 0 2 x