Lecture 31 INTEGRATION

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1 Lecture 3 INTEGRATION Substitution. Example. x (let u = x 3 +5 x3 +5 du =3x = 3x 3 x 3 +5 = du 3 u du =3x ) = 3 u du = 3 u = 3 u = 3 x3 +5+C. Example. du (let u =3x +5 3x+5 = 3 3 3x+5 =3 du =3.) = 3 du u = 3 ln u = 3 ln(3x +5)+C. Example 3. x +x (let u =+x du = +x x udu = = u du = u 3 =x du =x) = u 3 3 = (+x ) +x 3.

2 Integration : Substitution (cont d). Example. Example. cos x (+sin x) = du u = u du = u = u = +sin x. Lecture 3 (let u = + sin x du = cos x du = cos x) x log e x (let u = log e x du = x du = x ) = log e x x = u du = log e u = log e (log e x)+c.

3 Lecture 33 Partial Fractions. Example. Express x (x )(x 3) = A x + B x 3 x (x )(x 3) in the form A x + B x 3. = A(x 3)+B(x ) (x )(x 3) x =A(x 3) + B(x ) Substitute x =3. B = 5. Substitute x = A =3 A = 3. Alternatively,(A + B)x (3A +B) = x A + B = & 3A +B = 3( B)+B = B =5 A = 5= 3. x (x )(x 3) = 3 x + 5 x 3. Note: The degree of numerators is less than the degree of the demominators for breaking up of partial fractions. Example. Decompose x+3 (x +)(x ) x+3 (x +)(x ) = Ax+B x + x = (Ax+B)(x )+C(x +) (x +)(x ) into partial fractions. (since the degree of the numerators are at most, one less than the degree of the denominators.) x +3=(Ax + B)(x ) (x +) 7=6C C = 7 6 When x =0, 3= B +C 3= B B = 3. When x =, 5= (A + B)+c(3) = (A 3 )+ 7 6 (3) A = 7 6 x+3 (x +)(x ) = 7 6 x 3 x x = 7 6x 7x+ 6x +. Example 3. Decompose (x )(x+). (x )(x+) = A x + B x+ (x+) =A(x +) + B(x )(x +)+C(x ) When x =, 9A = A =9. When x =, 3C = C = 3. When x =0, =A B C, B = A C = + 3 B = 9. (x )(x+) = 9 x + 9 x+ + 3 (x+) = 9x 8 9x+9 3(x+).

4 Integration with partial fractions. Example. Find (x )(x ) = A x + (x )(x ). B x A =. Hence (x )(x ) = x + x. (x )(x ) = ( x + x ) Lecture 34 =A(x ) + B(x ). If x =,B =. Ifx =, A = = ln x +ln x =ln x x

5 Lecture 35 Example. Find 3x +5x+7 (x )(x ). First, (x )(x ) = x 3x +, which has the same degree as the numerator. So divide: 3 x 3x + ) 3x +5x +7 3x 9x +6 4x + Now use partial fractions: 4x+ (x )(x ) = A =5 A = 5 & B =9. A x + B x = A(x )+B(x ) (x )(x ) 4x +=A(x ) + B(x ) So, ( ) 3x +5x+7 (x )(x ) = (x )(x ) = ( 3 5 x + 9 x ) =3x 5 ln x +9ln x =3x +ln (x )9 (x ) 5

6 Lecture 36 Integration by Parts. d d (uv) =u v + v d u u dv = d du (uv) v. u dv = uv v du i.e., udv= uv vdu. Example. Find xe x. Let u = x du =,v = ex xe x = xe x e x = xe x e x = e x (x ) Example. Find x cos x. x cos x= x d sin x = x sin x sin x d x = x sin x sin x = x sin x + cos x

7 Lecture 37 Integration by Parts (cont d) Example. Find log e x. Let u =lnx, v = x log e x= log e x = x ln x = x ln x x Example. Find sin x. Let u = sin x, v = x. sin x= sin x = x sin x x let w = x x = dw x = x sin x ( w x dw ) x = x sin x + w dw = x sin x + w = x sin x +( x )

8 Lecture 38 Trigonometric Integrals. Powers of sin x Example. sin x= cos x Example. sin x= ( cos x) = (x sin x)+c Example 3. sin 3 x= sin x sin x = ( cos x) sin x(let u = cos x du = sin x) = ( u ) du = (u 3 u3 )+C = 3 cos3 x cos x Example 4. sin 4 x= (sin x) = ( ( cos x)) = 4 ( cos x + cos x) = 4 ( cos x + ( + cos 4x)) = 4 x sin x + (x + 4 sin 4x)) = 3 8 x 4 sin x + 3 sin 4x Example 5. sin 5 x= sin 4 x sin x = ( cos x) sin x(let u = cos x du = sin x) = ( u ) du = ( u + u 4 ) du = (u + 3 u3 + u5 5 )+C = 3 cos3 x cos5 x 5 cos x For sin n x, for odd powers, use the substitution method u = cos x. For even powers, use the fact that sin x = ( cos x)

9 tan n xfor n = Z + Lecture 39 Example. tan x= sin x cos x = ln cos x =ln cos x = ln sec x Example. tan x= (sec x ) = tan x x Example 3. tan 3 x= tan x(sec x ) = tan x sec x tan x = tan xd(tan x) tan x = tan x + ln cos x Example 4. tan 4 x= tan x(sec x ) = tan x sec x tan x = tan xd(tan x) tan x = 3 tan3 x (tan x x)+c = x + 3 tan3 x tan x Example 5. tan 5 x= tan 3 x(sec x ) = tan 3 x sec x tan 3 x = tan 3 xd(tan x) tan 3 x = 4 tan4 x ( tan x + ln cos x)+c = 4 tan4 x tan x ln cos x

10 Lecture 40 Trigonometric integrals Example. sin x cos x(u = cos x, du = sin x) = udu = u = cos x Alternatively, sin x cos x(v = sin x, dv = cos x) = vdv = v + K = sin x + K sin Note: x and cos x differ only by a constant, so although the answers look different, they are really the same, only expressed differently, because C and K are arbitrary constants. Example. sin / x cos x(u = sin x du = cos x) = u / du = u3/ 3 = 3 sin3/ x Example 3. cos 7 x sin 4 x = cos 6 x sin 4 x cos x(split the odd power) = ( sin x) 3 sin 4 cos x(u = sin x du = cos x) = ( u ) 3 u 4 du = ( 3u +3u 4 u 6 )u 4 du = (u 4 3u 6 +3u 8 u 0 )du = u5 5 3u u9 9 u = 5 sin5 x 3 7 sin7 x + 3 sin9 x sin x

11 Example 4. cos 7 x (u = sin x u = cos x) sin 4 x = cos 6 x cos x sin 4 x = ( sin x) 3 cos x sin 4 x = ( u ) 3 du u 4 = 3u +3u 4 u 6 u du 4 = (u 4 3u +3 u )du = u u +3u 3 u3 = 3 (sin x) 3 + 3(sin x) + 3 sin x 3 sin3 x

12 Lecture 4 Integration using tangents of half angles Let t = tan x. Then sin x = t dt = sec x dt = ( + tan x ) dt = ( + t ) dt = +t = dt +t Example. ( 3+ cos x = 3+ = +t, cos x = t +t, tan x = t t. ) dt +t ( ) t +t ( dt ) +t ( 3(+t )+( t )) +t = dt 3+3t + t = dt 5+t = 5 tan t 5 = 5 tan tan x 5 Example. cos x 3+ cos x = (3+ cos x) 3 3+ cos x = 3 3+ cos x = x 3 5 tan tan x 5 = x 3 5 tan tan x 5

13 Lecture 4 Integrals with quadratic denominators Example. Example. Example 3. Example 4. Example x = 4( 9 4 +x ) = x = 4 ( 3 ) tan x 3 = 6 tan x 3 = 4x 9 x +4x+ = x 9 4 = ln (x + (x+) 3 = du u 3 = 3 = 3 3x +x+ = 3 = 3 = 3 = 3 = 3 x 9 4 u ln 3 x+ ln 3 u+ 3 x++ 3 x + 3 x+ 3 x + 3 x (x+ 3 ) + 9 du u + 9 ( ) 3 tan ) (Let u = x + du = ) 3u Let u = x + 3 = tan 3(x + ) = 3x+ tan +4x x = (x 4x) = 5 (x 4x+4) = 5 (x ) = du 5u = 5 ln 5+u 5 u = 5 ln 5 +x 5+ x du = (Let u = x du = )

14 Example 6. x x +6x+ = (x+6) 6 x +6x+ = x+6 x +6x+ 6 x +6x+ = ln(x +6x +) 3 x +6x+ = ln(x +6x +) 3 x +6x+9 8 = ln(x +6x +) 3 (x+3) 8 = ln(x +6x +) 3 du u 8 ( = ln(x +6x +) 3 ln 8 = ln(x +6x +) 3 8 u 8 u+ 8 x+3 ln 8 Let u = x +3 du = ) x+3+ 8

15 Lecture 43 Substitutions using trigonometry. - for a x, a + x, and x a, use substitutions x = a sin θ, x = a tan θ and x = a sec θ respectively. Example. x 9 x (Let x = 3 sin θ = 3 cos θdθ) = 9 sin θ3 cos θ dθ 9 9 sin θ = 7 sin θ cos θ dθ 3 cos θ =9 sin θdθ = 9 ( cos θ) dθ = 9 [θ sin θ]+c where x = 3 sin θ, sin θ = x 3 θ = sin x 3 = 9 (θ sin θ cos θ)+c = 9 (θ sin θ cos θ)+c = 9 (sin x 3 x 9 x 3 3 )+C = 9 (sin x 3 x 9 9 x )+C

16 Lecture 44 Reduction Formulae Example. If I n = sin n x(n is an integer). Express I n in terms of I n and hence evaluate π/ 0 sin 5 x(application of Integration by Parts). Solution. I n = sin n x = sin n x sin x(let u = sin n x and v = sin x u =(n ) sin n x. cos x & v = cos x) = cos x sin n x + (n ) sin n cos x cos x = cos x sin n x +(n ) sin n cos x = cos x sin n x +(n ) sin n ( sin x) = cos x sin n x +(n ) sin n x (n ) sin n x = cos x sin n x +(n )I n (n )I n (n )I n + I n = cos x sin n x(n )I n I n (n +)= cos x sin n x +(n )I n I n = n cos x sinn x + n n I n I 5 = π/ 0 sin 5 x =[ 5 cos x sin4 x I 3] π/ 0 =[ 5 cos x sin4 x ( 3 cos x sin x + 3 I )] π/ =[ 5 cos x sin4 x ( 3 cos x sin x + 3 =[ 5 cos x sin4 x 4 5 cos x sin x sin x)] π/ sin x] π/ 0 =[ 5 cos x sin4 x 4 5 cos x sin x 8 5 cos x]π/ 0 = ( 8 5 cos 0) = Example. If I n = (log e x) n show that I n = (log e ) n ni n hence evaluate (log e x) 4.

17 I n = (log e x) n (Let u = (log e x) n & v = = x(log e x) n n u = n x (log e x) n,v = x) x (log e x) n x =[x(log e x) n ] n (log e x) n = (ln ) n ni n I 4 = (log e x) n = (ln ) 4 4I 3 = (ln ) 4 4((ln ) 3 3I ) = (ln ) 4 8(ln ) 3 +I = (ln ) 4 8(ln ) 3 + ((ln ) I ) = (ln ) 4 8(ln ) 3 + 4(ln ) 4I...( ) = (ln ) 4 8(ln ) 3 + 4(ln ) 4 ln x(let u = log x, v = = (ln ) 4 8(ln ) 3 + 4(ln ) 4[[x ln x] ] = (ln ) 4 8(ln ) 3 + 4(ln ) 4[x ln x x] = (ln ) 4 8(ln ) 3 + 4(ln ) 4( ln +) = (ln ) 4 8(ln ) 3 + 4(ln ) 48 ln + 4 ( ) Also I = (ln ) I 0 = (ln ) = (ln ) [x] = ln. u = x,v = x) Example 3. Let I n = sec n x. Show that I n = tan x secn x n + n n I n d Note: sec x = d (cos x) = (cos) ( sin x) = sin x cos x = tan x sec x. I n = sec n x sec x[let u = sec n xv = sec x u =(n )(sec x) n 3 tan x sec x =(n )(sec x) n tan x, v = tan x] = tan x sec n x (n ) tan x(sec x) n = tan x sec n x (n ) (sec) n tan x = tan x sec n x (n ) tan x(sec x) n = tan x sec n x (n ) (sec x ) sec n x = tan x sec n x (n ) sec n x+(n ) sec n x So sec n x+(n ) sec n x=(n ) sec n x = tan x sec n x +(n ) sec n x= tan x sec n x +(n )I n So sec n x= tan x secn x n + n n I n

18 Example 4. Let I n = tan n x. Then: I n = tan n x tan x = tan n x(sec x ) = tan n x sec x tan n x = tan n x sec x I n [Let u = tan x du = sec x] = u n I n = un n I n = tann x n I n