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12 x1 t05 05 integration with inverse trig (2013)

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Nigel Simmons
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Integration Involving Inverse Trig
Integration Involving Inverse Trig  sin 1    c dx x  a x 2 2  a
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c  
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. dx i  4  x2
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. dx i  4  x2 1  x   sin    c 2
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. dx i  ii  dx 4  x2 9  x2 1  x   sin    c 2
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. dx i  ii  dx 4  x2 9  x2 1  x  1 1  x   sin    c  tan    c 2 3 3
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. 1 dx dx i  ii  dx iii  4  x2 9 x 2 0 2  x2 1  x  1 1  x   sin    c  tan    c 2 3 3
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. 1  x   sin 1   1 dx dx i  ii  dx iii    4  x2 9 x 2 0 2 x 2   2  0 1  x  1 1  x   sin    c  tan    c 2 3 3
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. 1  x   sin 1   1 dx dx i  ii  dx iii    4  x2 9 x 2 0 2 x 2   2  0 1 1 1  x  1 1  x   sin  sin 1 0  sin    c  tan    c 2 2 3 3
Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a OR  cos   a 1 dx 1 1  x   a 2  x 2  a tan  a   c   e.g. 1  x   sin 1   1 dx dx i  ii  dx iii   4  x2 9 x 2 0 2 x 2   2  0 1 1 1  x  1 1  x   sin  sin 1 0  sin    c  tan    c 2 2 3 3   0 4   4
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   dx 1 e 2 x
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   dx 1 e 2 x dx  sin e log 2 0 ex   1 x 0  log 2 1  e 2x
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   dx 1 e 2 x dx  sin e log 2 0 ex   1 x 0  log 2 1  e 2x  sin 1 e 0  sin 1 e log 2 1 1 1  sin 1  sin 2     2 6   3
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx dx 1 e 2 x 4  9x2 dx  sin e log 2 0 ex   1 x 0  log 2 1  e 2x  sin 1 e 0  sin 1 e log 2 1 1 1  sin 1  sin 2     2 6   3
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx  dx dx 1 e 2 x 4  9x 2 4 2 3 x dx  sin e log 2 0 ex   1 x 0 9  log 2 1  e 2x  sin 1 e 0  sin 1 e log 2 1 1 1  sin 1  sin 2     2 6   3
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx  dx dx 1 e 2 x 4  9x 2 4 2 3 x dx  sin e log 2 0 ex   1 x 0 9  log 2 1  e 2x 1 dx    sin 1 e 0  sin 1 e log 2 3 4 2 x 1 1 1 9  sin 1  sin 2     2 6   3
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx  dx dx 1 e 2 x 4  9x 2 4 2 3 x dx  sin e log 2 0 ex   1 x 0 9  log 2 1  e 2x 1 dx    sin 1 e 0  sin 1 e log 2 3 4 2 x 1 1 1 9  sin 1  sin 2 1 1  3 x   sin    c   3 2   2 6   3
0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx  dx dx 1 e 2 x 4  9x 2 4 2 3 x dx  sin e log 2 0 ex   1 x 0 9  log 2 1  e 2x 1 dx    sin 1 e 0  sin 1 e log 2 3 4 2 x 1 1 1 9  sin 1  sin 2 1 1  3 x   sin    c   3 2   2 6   3 Exercise 1E; 2, 3, 4a, 5b, 6 & 7bdf, 9, 12, 13, 16, 17, 20, 22

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12 x1 t05 05 integration with inverse trig (2013)

  • 1. Integration Involving Inverse Trig
  • 2. Integration Involving Inverse Trig  sin 1    c dx x  a x 2 2  a
  • 3. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a
  • 4. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c  
  • 5. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. dx i  4  x2
  • 6. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. dx i  4  x2 1  x   sin    c 2
  • 7. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. dx i  ii  dx 4  x2 9  x2 1  x   sin    c 2
  • 8. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. dx i  ii  dx 4  x2 9  x2 1  x  1 1  x   sin    c  tan    c 2 3 3
  • 9. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. 1 dx dx i  ii  dx iii  4  x2 9 x 2 0 2  x2 1  x  1 1  x   sin    c  tan    c 2 3 3
  • 10. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. 1  x   sin 1   1 dx dx i  ii  dx iii    4  x2 9 x 2 0 2 x 2   2  0 1  x  1 1  x   sin    c  tan    c 2 3 3
  • 11. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a 1 OR  cos   a dx 1 1  x   a 2  x 2  a tan  a   c   e.g. 1  x   sin 1   1 dx dx i  ii  dx iii    4  x2 9 x 2 0 2 x 2   2  0 1 1 1  x  1 1  x   sin  sin 1 0  sin    c  tan    c 2 2 3 3
  • 12. Integration Involving Inverse Trig  sin 1    c dx x  xc  a x 2 2  a OR  cos   a 1 dx 1 1  x   a 2  x 2  a tan  a   c   e.g. 1  x   sin 1   1 dx dx i  ii  dx iii   4  x2 9 x 2 0 2 x 2   2  0 1 1 1  x  1 1  x   sin  sin 1 0  sin    c  tan    c 2 2 3 3   0 4   4
  • 13. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx
  • 14. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   dx 1 e 2 x
  • 15. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   dx 1 e 2 x dx  sin e log 2 0 ex   1 x 0  log 2 1  e 2x
  • 16. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   dx 1 e 2 x dx  sin e log 2 0 ex   1 x 0  log 2 1  e 2x  sin 1 e 0  sin 1 e log 2 1 1 1  sin 1  sin 2     2 6   3
  • 17. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx dx 1 e 2 x 4  9x2 dx  sin e log 2 0 ex   1 x 0  log 2 1  e 2x  sin 1 e 0  sin 1 e log 2 1 1 1  sin 1  sin 2     2 6   3
  • 18. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx  dx dx 1 e 2 x 4  9x 2 4 2 3 x dx  sin e log 2 0 ex   1 x 0 9  log 2 1  e 2x  sin 1 e 0  sin 1 e log 2 1 1 1  sin 1  sin 2     2 6   3
  • 19. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx  dx dx 1 e 2 x 4  9x 2 4 2 3 x dx  sin e log 2 0 ex   1 x 0 9  log 2 1  e 2x 1 dx    sin 1 e 0  sin 1 e log 2 3 4 2 x 1 1 1 9  sin 1  sin 2     2 6   3
  • 20. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx  dx dx 1 e 2 x 4  9x 2 4 2 3 x dx  sin e log 2 0 ex   1 x 0 9  log 2 1  e 2x 1 dx    sin 1 e 0  sin 1 e log 2 3 4 2 x 1 1 1 9  sin 1  sin 2 1 1  3 x   sin    c   3 2   2 6   3
  • 21. 0 iv  Find sin 1 e x and hence evaluate  d ex dx  log 2 1  e 2x dx ex d sin 1 e x   v  dx  dx dx 1 e 2 x 4  9x 2 4 2 3 x dx  sin e log 2 0 ex   1 x 0 9  log 2 1  e 2x 1 dx    sin 1 e 0  sin 1 e log 2 3 4 2 x 1 1 1 9  sin 1  sin 2 1 1  3 x   sin    c   3 2   2 6   3 Exercise 1E; 2, 3, 4a, 5b, 6 & 7bdf, 9, 12, 13, 16, 17, 20, 22