SUBSTITUTION I.. f(ax + b)
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1 Integrtion SUBSTITUTION I.. f(x + b) Grhm S McDonld nd Silvi C Dll A Tutoril Module for prctising the integrtion of expressions of the form f(x + b) Tble of contents Begin Tutoril c 004 g.s.mcdonld@slford.c.uk
2 . Theory. Exercises 3. Answers 4. Stndrd integrls 5. Tips Full worked solutions Tble of contents
3 Section : Theory 3. Theory Consider n integrl of the form f(x + b)dx where nd b re constnts. We hve here n unspecified function f of liner function of x Letting u = x + b then du dx =, nd this gives dx = du This llows us to chnge the integrtion vrible from x to u f(x + b)dx = f(u) du
4 Section : Theory 4 The finl result is where u = x + b f(x + b)dx = f(u) du This is generl result for integrting functions of liner function of x Ech ppliction of this result involves dividing by the coefficient of x nd then integrting
5 Section : Exercises 5. Exercises Click on EXERCISE links for full worked solutions (0 exercises in totl). Perform the following integrtions: Exercise. (x ) 3 dx Exercise. cos(3x + 5)dx Exercise 3. e 5x+ dx Theory Stndrd integrls Answers Tips
6 Section : Exercises 6 Exercise 4. sinh 3x dx Exercise 5. dx x Exercise 6. dx + (5x) Exercise 7. sec (7x + )dx Theory Stndrd integrls Answers Tips
7 Section : Exercises 7 Exercise 8. sin(3x )dx Exercise 9. cosh( + x)dx Exercise 0. tn(9x )dx Theory Stndrd integrls Answers Tips
8 Section 3: Answers 8 3. Answers. 8 (x )4 + C, sin(3x + 5) + C, e5x+ + C, cosh 3x + C, ln x + C, 6. 5 tn 5x + C, tn(7x + ) + C, cos(3x ) + C, 9. sinh( + x) + C, 0. 9 ln cos(9x ) + C.
9 Section 4: Stndrd integrls 9 4. Stndrd integrls f (x) x n x f(x)dx f (x) f(x)dx xn+ n+ (n ) [g (x)] n g (x) ln x g (x) g(x) [g(x)] n+ n+ (n ) ln g (x) e x e x x x ln ( > 0) sin x cos x sinh x cosh x cos x sin x cosh x sinh x tn x ln cos x tnh x ln cosh x cosec x ln tn x cosech x ln tnh x sec x ln sec x + tn x sech x tn e x sec x tn x sech x tnh x cot x ln sin x coth x ln sinh x sin x cos x x x sin x 4 sinh x sinh x 4 x + sin x 4 cosh x sinh x 4 + x
10 Section 4: Stndrd integrls 0 f (x) f (x) dx f (x) f (x) dx +x tn x x ln +x x (0< x <) ( > 0) x ln x x+ ( x > >0) x sin x +x ( < x < ) x ln ln x+ +x x+ x ( > 0) (x>>0) x [ sin ( ) x +x ] x + x x [ [ sinh ( x cosh ( x ) + x +x ] ) + x ] x
11 Section 5: Tips 5. Tips STANDARD INTEGRALS re provided. Do not forget to use these tbles when you need to When looking t the THEORY, STANDARD INTEGRALS, AN- SWERS or TIPS pges, use the Bck button (t the bottom of the pge) to return to the exercises Use the solutions intelligently. For exmple, they cn help you get strted on n exercise, or they cn llow you to check whether your intermedite results re correct Try to mke less use of the full solutions s you work your wy through the Tutoril
12 Solutions to exercises Full worked solutions Exercise. (x ) 3 dx Let u = x then du dx = nd dx = du (x ) 3 dx = u 3 du = u 3 du = 4 u4 + C = 8 (x )4 + C. Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = nd f(u)du is u 3 du. Return to Exercise
13 Solutions to exercises 3 Exercise. cos(3x + 5)dx Let u = 3x + 5 then du dx = 3 nd dx = du 3 cos(3x + 5)dx = cos u du = 3 3 cos u du = 3 sin u + C = sin(3x + 5) + C. 3 Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = 3 nd f(u)du is cos udu. Return to Exercise
14 Solutions to exercises 4 Exercise 3. e 5x+ dx Let u = 5x + then du dx = 5 nd dx = du 5 e 5x+ dx = e u du = 5 5 e u du = 5 eu + C = 5 e5x+ + C. Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = 5 nd f(u)du is e u du. Return to Exercise 3
15 Solutions to exercises 5 Exercise 4. sinh 3x dx Let u = 3x then du dx = 3 nd dx = du 3 sinh 3x dx = sinh u du 3 = 3 sinh u du = 3 cosh u + C = cosh 3x + C. 3 Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = 3 nd f(u)du is sinh u du. Return to Exercise 4
16 Solutions to exercises 6 Exercise 5. dx x Let u = x then du dx = nd dx = du dx = x du u = = ln u + C du u = ln x + C. Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = nd f(u)du is du u. Return to Exercise 5
17 Solutions to exercises 7 Exercise 6. dx + (5x) Let u = 5x then du dx = 5 nd dx = du 5 dx + (5x) = du + u = 5 5 du + u = 5 tn u + C = = 5 tn 5x + C. Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = 5 nd f(u)du is du +u du. Return to Exercise 6
18 Solutions to exercises 8 Exercise 7. sec (7x + )dx Let u = 7x + then du dx = 7 nd dx = du 7 sec (7x + )dx = sec u du 7 = 7 tn u + C = tn(7x + ) + C. 7 Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = 7 nd f(u)du is sec udu. Return to Exercise 7
19 Solutions to exercises 9 Exercise 8. sin(3x )dx Let u = 3x then du dx = 3 nd dx = du 3 sin(3x )dx = sin u du 3 = 3 cos u + C = cos(3x ) + C. 3 Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = 3 nd f(u)du is sin udu. Return to Exercise 8
20 Solutions to exercises 0 Exercise 9. cosh( + x)dx Let u = + x then du dx = nd dx = du cosh( + x)dx = cosh u du = sinh u + C = sinh( + x) + C. Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = nd f(u)du is cosh u du. Return to Exercise 9
21 Solutions to exercises Exercise 0. tn(9x )dx Let u = 9x then du dx = 9 nd dx = du 9 tn(9x ) dx = tn u du = 9 9 tn u du = ln cos u + C 9 = ln cos(9x ) + C. 9 Note. The finl result cn lso be obtined using the generl pttern: f(x + b) dx = f(u) du where = 9 nd f(u)du is tn u du. Return to Exercise 0
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