Transcription of -Substitution
1 Joe Fosteru-SubstitutionRecall the substitution rule from MATH 141 (see page 241 in the textbook).TheoremIfu=g(x) is a differentiable function whose range is an intervalIandfis continuous onI, then f(g(x))g (x)dx= f(u) method of integration is helpful in reversing the chainrule (Can you see why?) Let s look at some 1 Find sec2(5x+ 1) 5x+ 1du= 5dx sec2(5x+ 1) 5dx= sec2(u)du= tan(u) +C= tan(5x+ 1) +CRemember, for indefinite integrals your answer should be in terms of the same variable as you start with, so remember tosubstitute back in 2 Evaluate the integral 532x 3 x2 3x+ 3x+ 1du= 2x 3dxhiu= (3)2 3(3) + 1 = 1u= (5)2 3(5) + 1 = 11 532x 3 x2 3x+ 1dx= 1111 udu= 111u 1/2du= 2u1/2 111= 2 11 2 1= 2( 11 1)In the above we changed the limits of integration to coincidewith our functionu.
2 Doing this means that we don t haveto substitute in foruat the end like in the indefinite integral in Example 1. But if you did substitute back and use theoriginal limits don t worry,you get the same answer. Try it for yourself now to 1 of 5 MATH 142 -u-SubstitutionJoe FosterExample 3 Find 1 8x x2= (x2 8x)= ((x 4)2 42)= 42 (x 4)2hiu=x 4du=dx 1 8x x2dx= 1 42 (x 4)2dx= 1 42 (u)2du= sin 1(u4)+C= sin 1(x 44)+CExample 3 illustrates that there may not be an immediately obvious substitution. In the cases that fractions and poly-nomials, look at the power on the numerator. In Example 3 we had 1, so the degree was zero. To make a successfulsubstitution, we would needuto be a degree 1 polynomial (0 + 1 = 1). Obviously the polynomial on the denominatorwas degree 2. So we forced a degree 1 polynomial to appear bycompleting the 2 of 5 MATH 142 -u-SubstitutionJoe FosterPractice ProblemsTry some of the problems below.
3 If you get stuck, don t worry!There are hints on the next page! But do try withoutlooking at them first, chances are you won t get hints on your 1 13x2 x3+ 5dx2. x3(2 +x4)5dx3. 70 4 + 3x dx4. 1(1 6t)4dt5. 0xcos(x2)dx6. sec (1/x)x2dx7. 1/21/6csc( t) cot( t)dt8. x2(x3+ 5)9dx9. 10xe x2dx10. (3t+ 2) /20cos(x) sin(sin(x))dx12. x(x2+ 1)2dx13. sin 1(x) 1 x2dx14. exsin(ex)dx15. 0 18x(4x2+ 1)2dx16. xx2+ 1dx17. 10 12x2(4x3 1)3dx18. sec(2 ) tan(2 )d 19. 2 16x(x2 1)2dx20. xsin(1 +x3/2)dx21. 1024x(4x2+ 4)2dx22. (1 + tan( ))5sec2( )d 23. 0 3 8x(2x2+ 3)2dx24. ecos(t)sin(t)dt25. 1016x(4x2+ 4)2dx26. tan 1(x)1 +x2dx27. 0 118x2(3x3+ 3)2dx28. sin(ln(x))xdx29. 10 8x(4x2+ 2)2dx30. exex+ 1dx31. cos( /x)x2dx32. sin(x)1 + cos2(x)dx33. 1cos2(t) 1 + tan(t)dtChallenge ProblemsBelow are some harder problems that require a little more thinking/algebraic manipulation to make the 10x x+ 1dx2.
4 12x2 12x+ 26dx3. x1 +x4dx4. (x+ 3) x 1dx5. x2 1 xdx6. x3 x2+ 1dx7. 1 21 4x x2dx8. /2 /2x2sin(x)1 +x6dx9. 3x 1x2+ 10x+ 28dx10. 40x 1 + 2xdx11. 1 1sin(x)1 + 1ex+ 1dxPage 3 of 5 MATH 142 -u-SubstitutionJoe FosterHints to Practice + 2 + 4 + 1 1 + 3t+ sin(x) + sin 1(x) 4x2+ + 4x3 2 1 +x3 4x2+ 1 + tan( ) 2x2+ cos(t) 4x2+ tan 1(x) 3x3+ ln(x) 4x2+ + cos(x) 1 + tan(t)Hints to Challenge 12. Complete the + 4 =x+ 1 17. Complete the This is an odd Complete the square,x=u (u 1)11. This is an odd 1 =ex+ 1 ex,u=ex+ 1 Page 4 of 5 MATH 142 -u-SubstitutionJoe FosterAnswers to Practice Problems1. 4 6 (2 +x4)6+C3. (1 6t)3+C5. 06. ln 1 + sin(1/x)cos(1/x) + (x3+ 5)10+ (3t+ 2)2(27t3+ 54t2+ 36t+ 8)+C11. 1 cos(1)12. 12x2+ 2+ (sin 1(x))2+C14.
5 Cos(ex) +C15. (x2+ 1) +C17. (2 )2+C19. 2720. 32cos(13/2x)+ (tan( ) + 1)6+ ecos(t)+ (tan 1(x))2+C27. 1828. cos(ln(x)) +C29. 1330. ln(ex+ 1) +C31. 1 sin( x)+C32. tan 1(tan2(x/2)+C33. 2 1 + tan(t) +CAnswers to Challenge 2 1(x 32)+ 1(x2) + (x 1)3/2(3x+ 17) +C5. 215 1 x(3x2+ 4x+ 8)+ (x2+ 1)3/2(3x2 1) +C7. sin 1(x+ 25)+C8. (x2+ 10x+ 28) 16 33tan 1( 3(x+5)3)+ ln(ex+ 1) +CPage 5 of 5)
