Exercise 7.2

2(4 2) 1x x x+ + + 10. x - x 11. x x + , x > 0 12. 3 5 3( 1)x - x 13. 3 3(2 3 ) x x+ 14. 1 (log ) mx x , x > 0, 1≠m 15. 29 4 x - x 16. 2 3xe + 17. 2x x e 18. 1/21 -tan xe x+ 19. 2/2 1/1 x x e - e + 20. 2 2 2 2 x - x x - x e - e e e + 21. tan² (2x - 3) 22. sec² (7 - 4 x) 23. 1/2 sin - x - x 24. 2cos 3sin 6cos 4sin x - x x x+ 25. 2 2 cos (1 tan )x - x 26. cos x x 27. sin 2 cos 2x x 28. cos 1 sin x x+ 29. cot x log sin x 30. sin 1 cos x x+ 31. ( ) sin 1 cos x x+ 32. 1 cot x+ 33. 1 tan- x 34. tan sin cos x x x 35. ( ) 1 log x x + 36. ( ) ( 1) logx x x x + + 37. ( ) 3 1 4sin tan -x x x8+ Choose the correct answer in Exercises 38 and 39. 38. 9/10 10 10 log 10 x e x x dx x + +∫ equals (A) 10x - x 10 + C (B) 10x + x10 + C (C) (10x - x 10)-1 + C (D) log (10x + x10) + C 39. 2 2 equals sin cos dx x x∫ (A) tan x + cot x + C (B) tan x - cot x + C (C) tan x cot x + C (D) tan x - cot 2 x + C 7.3.2 Integration using trigonometric identities When the integrand involves some trigonometric functions, we use some known identities to find the integral as illustrated through the following example. Example 7 Find (i) 2cos x dx∫ (ii) sin 2 cos 3x x dx∫ (iii) 3sin x dx∫ Solution (i) Recall the identity cos 2x = 2 cos² x - 1, which gives cos2x = 1 cos 2 x+ Therefore , = 1 (1 + cos 2 )2 x dx∫ = 1 1 cos 22 2dx x dx+∫ ∫ = 1 sin 2 C2 4 x x+ + (ii) Recall the identity sin x cos y = 1 2 [sin (x + y) + sin (x - y)] (Why?) Then = = 1 1 cos 5 cos C2 5- x x  + +   = 1 1cos 5 cos C10 2- x x + + (iii) From the identity sin 3x = 3 sin x - 4 sin 3 x, we find that sin3x = 3sin sin 3 x - x Therefore, 3sin x dx∫ = 3 1sin sin 34 4 x dx - x dx∫ ∫ = 3 1- cos cos 3 C4 12 x x+ + Alternatively, 3 2sin sin sinx dx x x dx=∫ ∫ = 2(1 - cos ) sinx x dx∫ Put cos x = t so that - sin x dx = dt Therefore, 3sin x dx∫ = ( ) 21 - t dt− ∫ = 2 C3 t- dt t dt - t+ = + +∫ ∫ = 31cos cos C3- x x + + Remark It can be shown using trigonometric identities that both answers are equivalent.

Search Any Question, Exercise or Concept Instantly