ntegration (p3) Integration by Inspection When differentiating using function of a function or the chain rule: If y = f(u), where in turn u = f(x)


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1 ntegrtion (p) Integrtion by Inspection When differentiting using function of function or the chin rule: If y f(u), where in turn u f( y y So, to differentite u where u +, we write ( + ) nd get ( + ) (. The result is product showing min function nd its derivtive or lmost its derivtive. An integrl of the form ( + ) contins min function nd lmost its derivtive nd to find the integrl we differentite the function which leds to the min function. Differentiting ( + ) gives ( + ) which is the function ( + ). we re trying to integrte but with n etr fctor of. Differentiting ( + ) divided by would give us the ect function we wish to integrte, ( + ) ( + ) ( ) so + ( + ) We hve compensted by dividing by nd simply reversed the process to get the integrl. Question Differentite nd compenste to get the ect integrl Answer ( + ) ( + ) sin cos ( + ) ( + ) ( + ) sin ( + ) ( + ) ( + ) sin cos ( + ) Compenste by: Dividing by Dividing by Dividing by Multiplying by nd dividing by ( + ) ( + ) sin ( + ) Simple stndrd integrls to lern. sin cos nd cos  sin nd e e nd sin cos e cos, cos sin, sin  e, e e sin cos
2 All the bove hve been done by inspection but there re some integrls which my hve to be done by substitution. Integrtion by Substitution The ide is to turn difficult integrl in terms of : into n esier one in terms of u Consider ( + ) Eliminte the min obstcle by letting u + Differentite to get the link between nd du: therefore Now substitute to completely eliminte ll trces of : u du u du u ( + ) To get the nswer bck in terms of the originl substitute bck for u: If there re limits then chnge them in ccordnce with the substitution: u + ( + ) becomes u du u 78. Now consider This cnnot be done by inspection. We cnnot compenste by dividing by ( + ) the vrible. This hs to be done by substitution. Let u +, du. ( u ) du The integrl becomes ( u) du ln u  u u u Now consider + substitution like this: Let tn θ, sec θ dθ The integrl becomes + tn θ The substitution u + sec θ, sec θ dθ θ (since sec θ + tn θ ) Logrithmic Integrtion would not work. We hve to turn to trig θ tn  d θ θ tn  Applying the chin rule to ln f(: f ( ln f( f ( f ( f ( f ( f ( ln f( The integrl of quotient, where the numertor is the differentil of the denomintor, is the nturl log of the denomintor. + ln ( + + ) + ln ( + )
3 + here, the numertor is not the ect differentil of the denomintor so we hve to do some + compensting gin. We need + in the numertor so put the in nd compenste with outside the integrl sign. ( + ) ln ( + + Prtil frctions leding to logrithmic integrtion. Rtionl functions with qudrtic denomintors re split into prtil frctions nd then integrted: ( + )( ) ( ) ( + ) ln () ln (+) ln ( ) ( + ) Integrtion by prts  used with mied products v When differentiting product: uv u dv du This cn be written u + v uv or + v nd the reverse: dv du u + v uv dv du u uv  v e v designte u so tht when it is differentited it becomes simpler: Let u, e e e  e. e  e cos sin  sin. sin os Sometimes fter squeezing out the second integrl needs Further squeezing out nd leves third integrl. v u nd re chosen nd the nswer found by tking out uv nd then performing second simpler integrtion. sin v u (cos  ( cos u v  v  cos + cos  cos + sin  sin v u u v v  cos + sin + cos Integrtion by prts is used to integrte the function ln by writing.ln nd letting u ln.ln ln.  ln  Further integrtion methods involve using the trigonometric identities.
4 Integrtion of odd powers of sin nd cos When differentiting sin n nd cos n we get n sin n cos nd n cos n sin respectively There is lwys single sin or cos tgged on so we prepre for this when integrting odd powers. sin sin sin Now use the identity cos + sin since (  cos sin sin cos cos sin  cos + cos cos sin. Note tht we compenste by dividing by. Similrly cos cos cos ( sin cos cos sin cos Integrtion of odd powers of sin nd cos sin  sin To integrte even powers of sin nd cos, we use the double ngle formul: cos sin cos  Chnging the subject of these two identities gives sin cos nd cos cos So sin And cos cos (  cos ( sin ) sin  most of the integrtion methods hve been deduced by first considering differentition of functions, it might be worth noting the following differentils. sec n n sec n sec tn n sec n tn sec n n sec tn n tn n n tn n sec tn n sec n+ tn n + Sin A + sin B sin Integrtion of products of sin nd cos using the fctor formul A + B cos A B is the first of the fctor formule Let A P + Q nd B P Q, this gives the formul sin(p+q) + sin(pq) sin P cos Q The other three formule re: sin(p+q) + sin(pq) sin P cos Q sin(p+q) + sin(pq) sin P cos Q sin(p+q) + sin(pq) sin P cos Q These formule enble us to integrte trigonometric products like sin cos. sin cos sin 8 + sin (using eqution bove) cos8 cos + 8
5 Finding the re under curve given prmetriclly The prmetric equtions of the circle centre (,) nd rdius r re: r cos θ, y r sin θ The re under the curve between nd r is written r y r r sinθ There is miture of θ nd here so substitution hs to be mde. Replce by θ θ d nd chnge the limits using θ cos r The integrl becomes r ( ) r θ θ θ r sinθ( r sinθ ) dθ r sin θ d θ r (  θ which is qurter of the re of circle! sin ) Integrtion checklist If the integrl is product of function nd its derivtive try inspection. If the integrl is product of function nd its derivtive nd inspection is out, try substitution. If the integrl is mied product eg. ( sin then go for integrtion by prts. If the integrl is frction with the derivtive bove function use logrithmic integrtion. Qudrtic denomintors cn be fctorised nd should led to prtil frctions. For even or odd powers of sin or cos use the pproprite trigonometric identities. 7 For products of sin nd cos use the fctor formul to substitute. 8 For ny other integrl you re on your own!
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