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Mathematics

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01_Straight_Line 02_Circle 03_Permutation_and_combination 04_Complex_numbers 05_Application_of_Derivative 06_Binomial_theorem 07_Progression__Series.pdf 08_Definite_Integral.pdf 09_Indefinite_Integral.pdf 10_Probability.pdf

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  1. INDEFINITE INTEGRAL DEFINITION Integration is the reverse process of differentiation. If F'(x) = f (x) then F(x) is antiderivative of the function f(x). 2. If F(x) is an antiderivative of f(x) then F(x) + c where c is a constant, is also antiderivative of f(x) thus function posses infinitely many antiderivatives. Symbolically we can represent it as ff(x) dx = F(x) + c. Hence we have d —IF(x) +c] f (x) ff(x) dx F(x) +c where c is the constant of integration. dx Rules of Indefinite Integration: fkf(x) dx kff(x) dx where k is a constant. i) ii) ffl(x) dx ± ff2(x) dx ± ff3(x) dx... ± ffn(x) dx . If ff(x) dx Example 1 : (iii) Solutions: (i) (iii) f(ax + b) = F(x) + c, then ff(ax +b) dx Evaluate ftan2 xdx . f3xl/2(1 + x3/2)dx. a (ii) feX(cosx - sinx)dx. (iv) fsin3xcos2xdx . ftan2 xdx sec2 x -1) dx = tan x —x + c. feX(cosx - sinx)dx. Here eX(cosx - sinx) is the derivative of ex cosx. f3xl/2(1 + x3/2)dx. Here + is the derivative of (1 + x ) 3/2 2 3/2 2 (iv) fsin3xcos2xdx . When solving such problems it is expedient to use the following trigonometric identities : sin(mx) cos(nx) = (1/2) [sin(m - n)x + sin(m + n)x] sin(mx) sin(nx) = (1/2) [cos(m - n)x - cos(m + n)x] cos(mx) cos(nx) = (1/2) [cos(m - n)x + cos(m + n)x] Here sin 3x cos 2x = (1/2) [sin (5x) + sin x] 1 1 cos5x = —f(sin5x+sinx)dx + COS X + C . 2 2 5 BASIC FORMULAE Antiderivatives or integrals of some of the widely used functions (integrands) are given below. dx n +1 d 1 —(lnlxl) dx x d dx f xn dx f— dx f ex dx Inlx14-c
  2. 2 d —(aX) - (ax Ina) dx d — (sin x) = cosx dx d — (cos x) —sinx dx d —(tanx) = sec2 x dx d Indefinite Integral ax fax dx Ina Icos x dx = sinx+c fsinxdx -cosx + c f sec2 x dx = tan x +c — (cosec x) = (—cotx cosec x) fcosecx cot x dx cos ecx + c dx d — (sec x) = sec x tan x dx d cosec2x — (cot x) dx d dx d tan dx d — (sec dx f sec x tan x dx = secx +c fcosec2x dx cotx + c 1 x dx = Sin a x a x a x) 1 a 2 1 2 cosx dx 1 1 = —tan 1 — + C a a dx = sec-I (x) +c fcot x dx f tan x dx f sec x dx dx = In I sinx14-c sinx —sinx dx —In Icosx I+C or Inl secxl+c cos x sec x(sec x +tan x) dx = In I secx + tanx I +c sec x + tan x or In tan +c cos ecx(cot x — cos ecx) f cos ecxdx f dx cot x — cos ecx or or In tan— + c 2 STANDARD SUBSTITUTIONS - Inlcot x cosec xl +c For terms of the form + a2 or put x = a tano or a coto For terms of the form — a2 or —a2 , put x = a sec 0 or a cosec0 For terms of the form a2 — or A.FF, put x = a sin 0 or a cos0 If both are present, then put x = a cos0. For the type (x— a) (b —x) , put x = a cos20 + b sin20 For the type +a2 ± x or x ± —a put the expression within the bracket = t. 1 1 For the type (x + a) (x + b) 1 For , m, rue N 1 1 (n e N, n >1), put or (and >1), again put (X+a) = t (x + b)
  3. Indefinite Integral STANDARD FORMULAE x-a dx dx dx dx 1 2a x a—x u2 +a2du u -a2 Example 2 : Solution: Evaluate dx (x dx (x dx —Inu+ u2 —a2 — sin 3)4/5 3)4/5 dx (x dx 4/5 1/5 x) put Hence I = — 4 (ii) f (x (x -3)4/5 dx dx 5 4 Put x = acos20 + bsin20, the given integral becomes. 2(b —a) sin Ocos Od0 {(a cos2 0 + bsin2 0 —a) 2 (b — a)sin Ocos Od0 (b —a)sin0cos0 b—a 12 do = 20 + c = b—a (b —a cos2 0 —b sin2 OF 2sin x b a a
  4. 4 METHODS OF INTEGERATION Indefinite Integral Form 1: dx dx or ax2 +bx +c dx . ax2 -k + C ax2 + + c Working Rule: Express the integral in the form dx 2 expressing ax2 + bx + c as sum or difference of two perfect squares i.e. ax by Form 2: 2 b 2 c 4a2 a dx or f Q(x) is of one degree more than P(x). Working Rule: Express P(x) = + where X, are constants. Then evaluate and by equating the dx coefficients of xl and xo. Hence it deduces to some simplified form. Form 3: a) Integration of Rational Algebraic functions (P(x)/Q(x)): A function of the form P(x)/Q(x), where P(x) and Q(x) are polynomials, is called a rational function. If degree of P(x) < degree of Q(x). When Q(x) is expressible as the product of non-repeated linear factors. Q(x) = (x — — a) an) Working Rule: b) Express Q(x) (x —al) (x —a2) Evaluate Al, A2,... , An by comparing the coefficient of x on both sides. When Q(x) is expressible as the product of linear factors such that some of them are repeated i.e. Q(x) = (x — (x — al) .....(x— an) V k e N. Working Rule: c) Express Q(x) (x —a) (x —a)2 (X —a)k (X —al) (X —a) (X —an) When Q(x) contains quadratic factors which are non-repeating i.e. Q(x) = (ax2 + bx + c)(x — al) (x — a) ...(x— an) with condition on (b 2 — 4ac < O). Working Rule: (X —an) d) Express Q(x) ax2 +bx+c (x —al) (x —a) When Q(x) quadratic factors which are repeating i.e. Q(x) = — al). .. (x— an) with condition on (b2 — 4ac < 0). Working Rule: Express x) ax2 + -k c (a + + c) 2 (X —an)
  5. Indefinite Integral Form 4: dx 4 Working Rule: 5 2 a 2 x i) Divide numerator and denominator by we get f dx 2 x 2 a a 4 — dx = dt and therefore square to get +— ii) Substituting = t which gives 2 x x x Note: It is to be noted that even powers of x exist in such integrals. Form 5: dx dx dx asinx+bcosx +c a +bsinx a + bcos x Working Rule: 1 Replace sin x = where t = tan , cos x = 1 1 Form 6: (a sin x + b cos x)dx (c sin x +dcosx) Working Rule: d x 2 Express numerator = A(Denominator) + B (Denominator) where A, B are constants. Find the constants A, B by equating the coefficients of sin x and cos x on both sides. Form 7: (a sin x +bcosx + c)dx (psinx + qcosx + r) Working Rule: d Express numerator = A(Denominator) + B (Denominator) where A, B are constants. Form 8: dx dx 2 (asin2 x +bcos2 x +c) (asinx +bcosx) Working Rule: Divide numerator and denominator by cos2 x then replace sec2 x if any denominator by (1 + tan2 x) and then substitute tan x = t. Form 9: f sinm xcosn x dx V m, n e N Working Rule: Case I: If m is odd integer and n is even integer then put cos x = t. Case Il: If n is odd integer and m is even integer then put sin x = t. Case Ill: If m and n both are odd integers then put the term which have higher degree equal to t.
  6. 6 Case IV: If m and (1 + COS cos2 x 2 n both are natural numbers then use identities sin Indefinite Integral (1 - cos2x) x 2 and Case V: If m and n are rational numbers and Example 3 : Solution : Form 10: f sin3 x cos5 x dx Evaluate : f sin3 x cos5x dx Let cosx = p - sinx dx = dp fp5(1 —p2)dp — fp7dp —fp5dp 8 Alternatively: I (1 R2)2dR if sin x sin4 x fR3dR-f2R5 dR +fR7dR 4 2 6 R, 2sin 6 2 is l- (negative integer) then put tan x = t. cos8x cos6 x 8 6 cos dx x sin8 x 8 Integration by Parts. Working Rule: Suppose we are given two different functions f(x) and g(x) such that we have to find ff(x) .g(x) dx then d dx = dx dx dx dx where we have f(x) as first function and g(x) as second function. Rules for selecting f(x) and g(x): We know f(x) is first function and g(x) is second function. ILATE rule is applied for choosing first and second functions. I Inverse Function L Logarithmic Function A Algebraic Function T Trigonometric Function E Exponential Function Example 4 : Solution : Evaluate (i) flnx dx (iii) (i) flnxdx = Inx fl 2 x 2 2 2 (ii) f +a2dx dx-fldx = (x In x —x) +c x(2x) + a2 fldx f dx 2 2 2 2 a dx + 2 2 (iii)
  7. 7 2 1 dx + a dx a2 x2f1. dx = 2 + A—Sln x 2 a Tex [f(x) +f'(x)] dx Working Rule: Integration of such a function is done as follows: f ex [f(x) +f'(x)] dx eXf(x) +c . Form 12: feax sin(bx +c) dx or feax cos(bx +c) dx Working Rule: Indefinite Integral —x dx = x x 2 Form 11: 2 x 2 —x b a b a a 2 x 2 feax cos (bx + c) dx = cos(bx+c +)+k where tan feax sin(bx + c) dx = sin(bx+c +)+k where tan

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