MT1710 - 2018/19 : WEEK 7 Chapter III: Integration (continued) Section 3.2. - Techniques of integration (continued). Sub-section 3.2.2 - Substitution techniques. Examples 3.4 Integrate x (i) 2x +1 , (ii) x2 + 2x +1' x2 + 6x + 3 2x +3 x2+2x+10 . (iii) Workout: (i)After changing the given expression to 2 2x+1 x = 1 (1 - 2(2x+1) ), which can be achieved either by inspection, by partial fractions or by long division, we integrate in class by using 1 2t + 1 t dt = 1 2 1 (1 - 2 dt 2(2t + 1) (completed in class) (ii) Change the given expression according to, x2+6x+3 4x +2 x2+2x+1 = 1 + x2+2x+1 x x2+2x+1 + 4(x+1)-2 (x+1)2 . 4 2 =1+ (x + 1) (x+1)2 Then, integrate using the substitution x + 1 = u implicitly: Ï dt = t2 + 6t + 3 1 1 1 + t2 + 2t + 1 2 2 (t+ 1)2 ! dt (t + 1) 4 2 (x+ 1) + C, =x+ 4ln |x + 1| + for C a constant of integration.
2x+3 (iii) Change the given expression to x2+2x+10 x2+2x+10 2x+2 + (x+1)2+32 1 then integrate - t2 + 2t + 10 2t + 3 x dt = - t2 + 2t + 10 2t + 2 + (t + 1)2 + 32 1 ! dt. (completed in class) The following examples show substitutions that are more elaborated. Using integration by substitution, remember there are three steps to follow, *express the integrand function in the new variable, *replace the integration element, expressing it in the new variable, *replace the integral end limits to the values in the new variable. Examples 3.5 Evaluate the following integrals: 8 cos vt (i) 1 8 Vt dt, (ii) - V5+ t2 t(t2 + 1) dt . Workout: (i) the form of the integral suggests the use of the substitution t = u2. (completed in class) (ii) The factor (5 + t2) present in the denominator of the integrand function suggests the use of the substitution 5 + t2 = u2. (completed in class) Hwk: Try reworking this integral with the substitution t = v5 sinh u. Some standard substitutions: (a2 - [2) 1/2 (x2 - a2) 1/2 (x2 + a2) 1/2 (x2 + a2) is is is is x = a tan 0 x = a sin 0 x = a cosh 0 x = a sinh 0 (or x = a cos 0) (or x = a sec 0) (or x = a tan0) To understand these substitutions just use the formulae, sin2x + cos2x = 1, cosh2x - sinh2x = 1 and 1 + tan2x = 1/cos2x. N.B .- Every quadratic form may be rewritten in such standard form since 0 ax2 + 26x + c == {(ax + b)2 + ac -62}, a = 0
Several examples will now illustrate the use of quadratic forms. Examples 3.6 What are the substitutions appropriate to evaluate integrals with the expressions: