RAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IITJEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties


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1 M.Sc. (Mths), B.Ed, M.Phil (Mths) MATHEMATICS Mob. : M.Sc. (Mths), B.Ed, M.Phil (Mths) RAM RAJYA MORE, SIWAN XI th, XII th, TARGET IITJEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties CONTENTS Key Concept  I... EericiesI... EericiesII... EericiesIII... Solution Eercise Pge...
2 M.Sc. (Mths), B.Ed, M.Phil (Mths) Introduction THINGS TO REMEMBER Integrtion is clled the inverse process of differentition. This helps to find the function whose differentil coefficient is known. Thus, function f() is clled primitive or n ntiderivtive of function f(), if F () = f() Since, the differentil coefficient of constnt is zero. Hence, d d [F() + c] = F() + 0 = f() f() = F() + c where, is the nottion of integrtion, f() is clled the integrnd nd is clled the vrible of integrtion. c is the constnt of integertion nd cn tke ny constnt vlue. This shows tht F() nd F() + c re both integrls of the sme function f(). Thus, for different vlue of c, we obtin different integrl of f(). This implies tht the integrl of f() is not definite. Due to this F() is clled the indefinite integrl of f(). Stndrd Formul. n = n + c, (n ) n. = loge + c 3. e = e + c 4. = log e + c 5. sin = cos + c 6. cos = sin + c 7. sec = tn + c 8. cosec = cot + c 9. sec tn = sec + c 0. cosec cot = cosec + c. cot = log sin + c. tn = log cos + c 3. sec = log sec + tn + c = log tn 4 + c 4. cosec = log cosec cot + c = log tn + c
3 5. sin c 6. tn c 7. sec c 8. log c, 9. log c 0. log c sinh c. log c cosh c. sin c 3. log c 4. log c 5. e + b [f () + f () ] + = e + b f() + c e 6. e sin b = b ( sin b b cos b) + c e 7. e cos b = b Rules for Integrtion. { f() + g()} = f() + g(). k f() = k f(), where k is contnt. 3. d ( f() ) = f() 4. f() = g() + c g( b) f( + b) = c e b ( cos b + b sin b) + c = cosb tn c b 3 M.Sc. (Mths), B.Ed, M.Phil (Mths)
4 M.Sc. (Mths), B.Ed, M.Phil (Mths) Integrtion by Substitution Integrtion of certin function cnnot be btined directly, if they re not in one of the stndrd forms, but they my be reduced to stndrd forms by proper substitution. If g() is continuous diffecentible function, then to evlute integrls of the form I = f [g()]. g () We substitute g() = t nd g () = dt The substitution reduces the integrl to f(t)dt. After evluting this integrl we substitute bck the vlue of t. Function Substitution f( + ), f = sin or = cos f = sec or = cosec f = tn or = cot f, f b f, f b = cos = cos + b sin Integrtion of Different Types of Function. Integrtion of type sin m cos n () If m is n odd integer, put cos = t (b) If n is n odd integer, put sin = t (c) If m + n is negtive even integer, then put tn = t or cot = t. (d) If m nd re even nturl number, then converts higher power into higher ngles.. If the integrls re of the form b c, nd b c b, the epress + b + c s the sum of difference of two squres ie, in the form of perfect squre nd then, pply the stndrd results. p q p q 3. If the integrls re of the form, nd (p + q) b c b c b. Then put p + q = A (differentil coefficient of + b + c) + B. Find A nd B by compring the 4
5 M.Sc. (Mths), B.Ed, M.Phil (Mths) p q coefficients of nd constnt term on bothe the side of eqution. Then = b c b A + A b c. Now, we cn integrtes it esily. Similrly, other tow cses. b c 4. If the integrls re of the form ( b) l m nd ( b c), then put l + m = t. l m 5. If the integrl is of the form ( b) l m n then put + b = t nd for ( b) l, m put = t. 6. If the integrl is of the form ( b ) l m n b c then put t. l m n 7. If integrl is of the form denomintor (Dr) by nd, where k is ny constnt, then divide numertor (Nr) nd k = t. 8. If the integrl is of the form m n ( b) ( c d) b t. c d, where m nd n re positive integers, then put 9. If the intergrl re of the form cos b, nd b sin sin b sin cos c then multiply the Nr or Dr by sec nd then. put tn = t. (use sec = + tn ). cos, 0. If the integrls re of the form, nd cos bsin b cos sin sines nd cosines into their respective tngents of hlf the ngle nd then, put nd then convert tn t. p cos q sin r p cos q sin p cos. If the integrls re of the form,,, cos bsin n cos bsin cos bsin q sin nd then epress the numertor s cos bsin d Nr A(Dr) + B (Dr) + c [The vlue of c is zero in lst three forms] nd compre the coefficients of sin nd cos nd then, proceed. 5
6 M.Sc. (Mths), B.Ed, M.Phil (Mths) e be. If integrl is of the form then put e + be = A(ce + de ) + B(ce de ) ce de e Now, find ce Integrtion by Prts be de = A + B If u nd v re the differentible function fo, then u. v = u v d ( u) ( v). f ( ), where f() = ce + de. f ( ) ie, the integrl of the product of two fucntions = (first functions) (Integrl of the second function) Integrl of {(differentition of first function) ( Indtegrl of second function)}. How to Choose Ist nd IInd Function If two function re of different types tke the function s Ist which comes first in the word ILATE, where I stnds for inverse cirulr functio, L stnds for logrithmic fucntion, A stnds for lgebric function, T stnds for tirgonometric nd E for eponentil function. Some Importnt Points. If f() nd g() re sme functions, then {f() g()} is constnt.. As integrtion nd differentition re inverse process, mny times result of integrtion cn be verified by differentiting its perimitive of ntiderivtive. eg, log( ) log log ( ) c cn be verified by differentiting its perimitive. d log () log = log( ) log ( ) Some Integrls which cnnot be Found Any function continuous on n intervl (, b) hs n ntiderivtive n tht intervl In other words, there eists function F() such F () = f(). However, not every ntiderivtive F(), even when it eists, is epressibe in closed form in terms of elementry functions such s polynomils, trigonometric, logrithmic, eponentil functions etc. Then, we sy tht such ntiderivtives of integrls cnnot be fountd. 6
7 Note : If the integrl contins single logrithmic or single inverse trigonometric function tke unity s the second function. If the integrls of both the functions re known, the function which is esy to integrte is tken s the secon function. In certin cses integrtion by prts will led to simple eqution involving the integrl. Solve the eqution nd determine the integrl. 7 M.Sc. (Mths), B.Ed, M.Phil (Mths)
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