How to Integrate Trig Functions

Step-by-Step Guide

1. 1

   Step 1: Know the relevant trig identities.

   These trig identities may be useful to know.
   
   Pythagorean Identities (PI): sin2⁡(x)+cos2⁡(x)=1{\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1} sin2⁡(x)=1−cos2⁡(x){\displaystyle \sin ^{2}(x)=1-\cos ^{2}(x)} cos2⁡(x)=1−sin2⁡(x){\displaystyle \cos ^{2}(x)=1-\sin ^{2}(x)} sec2⁡(x)=tan2⁡(x)+1{\displaystyle \sec ^{2}(x)=\tan ^{2}(x)+1} csc2⁡(x)=cot2⁡(x)+1{\displaystyle \csc ^{2}(x)=\cot ^{2}(x)+1} Double Angle Identities (DAI): sin⁡(2x)=2sin⁡(x)cos⁡(x){\displaystyle \sin(2x)=2\sin(x)\cos(x)} cos⁡(2x)=cos2⁡(x)−sin2⁡(x){\displaystyle \cos(2x)=\cos ^{2}(x)-\sin ^{2}(x)} cos⁡(2x)=1−sin2⁡(x){\displaystyle \cos(2x)=1-\sin ^{2}(x)} cos2⁡(x)=12(1+cos⁡(2x)){\displaystyle \cos ^{2}(x)={\frac {1}{2}}(1+\cos(2x))} sin2⁡(x)=12(1−cos⁡(2x)){\displaystyle \sin ^{2}(x)={\frac {1}{2}}(1-\cos(2x))}
2. 2

   Step 2: Separate a sin2⁡(x){\displaystyle \sin ^{2}(x)}.

   ,,,, This is done to simplify the integrand.
   
   Let u=ln⁡(x){\displaystyle u=\ln(x)}.
   
   Then du=1xdx{\displaystyle du={\frac {1}{x}}dx}.
   
   This is done to simplify the integrand.
   
   Substitution yields ∫sin5⁡(u)du{\displaystyle \int \sin ^{5}(u)du} , The goal here is to get a term defined by a trig identity. ∫sin4⁡(u)sin⁡(u)du{\displaystyle \int \sin ^{4}(u)\sin(u)du} ∫(sin2⁡(u))2sin⁡(u)du{\displaystyle \int {(\sin ^{2}(u))}^{2}\sin(u)du} , Our goal here is to set up the integrand for another u-sub.
   
   Therefore, we want the integrand to consist of a trig function and it's known derivative. ∫(1−cos2⁡(u))2sin⁡(u)du{\displaystyle \int {(1-\cos ^{2}(u))}^{2}\sin(u)du} , Let w=cos⁡(u){\displaystyle w=\cos(u)}.
   
   Then −dw=sin⁡(u)du{\displaystyle
   -dw=\sin(u)du} Substitution yields −∫(1−w2)2dw{\displaystyle
   -\int {(1-w^{2})}^{2}dw} , Distribute the two binomials.
   
   Combine like terms. −∫1−2w2+w4dw{\displaystyle
   -\int 1-2w^{2}+w^{4}dw} , −+C{\displaystyle
   -+C} , w=cos⁡(u){\displaystyle w=\cos(u)} and u=ln⁡(x){\displaystyle u=\ln(x)} −+C{\displaystyle
   -+C} ,,,, This is done to simplify the integrand.
   
   Let u=7x{\displaystyle u=7x}.
   
   Then 17du=dx{\displaystyle {\frac {1}{7}}du=dx}.
   
   Substitution yields 17∫sin2⁡(u)cos2⁡(u)du{\displaystyle {\frac {1}{7}}\int \sin ^{2}(u)\cos ^{2}(u)du} , The goal here is to get the integrand in terms of one trig function. ∫du{\displaystyle \int {}du} , Simplify. 128∫(1−cos⁡(2u))(1+cos⁡(2u))du{\displaystyle {\frac {1}{28}}\int {(1-\cos(2u))(1+\cos(2u))}du} 128∫1−cos2⁡(2u)du{\displaystyle {\frac {1}{28}}\int {1-\cos ^{2}(2u)}du} , This is done to make the integrand integrable. 128∫1−(12(1+cos⁡(4u)))du{\displaystyle {\frac {1}{28}}\int {1-({\frac {1}{2}}(1+\cos(4u)))}du} , Distribute and combine like terms. 128∫12−12cos⁡(4u)du{\displaystyle {\frac {1}{28}}\int {{\frac {1}{2}}-{\frac {1}{2}}\cos(4u)}du} , u=7x{\displaystyle u=7x} 128+C{\displaystyle {\frac {1}{28}}+C} ,,,,, This is done to simplify the integrand.
   
   Let u=x2{\displaystyle u={\frac {x}{2}}}.
   
   Then 2du=dx{\displaystyle 2du=dx} Change the bounds of integration since the integral at hand is a definite integral.
   
   When x=0{\displaystyle x=0}, u=0{\displaystyle u=0} When x=π{\displaystyle x=\pi }, u=π2{\displaystyle u={\frac {\pi }{2}}} Substitution yields 2∫0π2cos4⁡(u)du{\displaystyle 2\int _{0}^{\frac {\pi }{2}}\cos ^{4}(u)du} , Get the integrand in to a form such that an identity can be used. 2∫0π2(cos2⁡(u))2du{\displaystyle 2\int _{0}^{\frac {\pi }{2}}{(\cos ^{2}(u))}^{2}du} ,, Simplify. 2∫0π212(1+cos⁡(2u)+cos2⁡(2u))du{\displaystyle 2\int _{0}^{\frac {\pi }{2}}{\frac {1}{2}}(1+\cos(2u)+\cos ^{2}(2u))du} , This is done to get cos2⁡(2u){\displaystyle \cos ^{2}(2u)} into a term we can integrate. 12∫0π21+2cos⁡(2u)+12(1+2cos⁡(4u))du{\displaystyle {\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}1+2\cos(2u)+{\frac {1}{2}}(1+2\cos(4u))du} , Simplify. 12∫0π232+2cos⁡(2u)+12cos⁡(4u)du{\displaystyle {\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}{\frac {3}{2}}+2\cos(2u)+{\frac {1}{2}}\cos(4u)du} , Since the bounds were converted, don't re-sub. 120π2{\displaystyle {\frac {1}{2}}{}_{0}^{\frac {\pi }{2}}} , =3π8{\displaystyle ={\frac {3\pi }{8}}} ,,,,, This is done to simplify the integrand.
   
   Let u=y{\displaystyle u={\sqrt {y}}}.
   
   Then 2du=1ydy{\displaystyle 2du={\frac {1}{\sqrt {y}}}dy}.
   
   Substitution yields 2∫tan4⁡(u)du{\displaystyle 2\int \tan ^{4}(u)du} , The goal here is to get a term we can replace with a Pythagorean Identity. 2∫tan2⁡(u)tan2⁡(u)du{\displaystyle 2\int \tan ^{2}(u)\tan ^{2}(u)du} , Only substitute one tan2⁡(u){\displaystyle \tan ^{2}(u)}.
   
   This is done to get a term and it's derivative in the integrand; this a set up for a u-sub. 2∫(sec2⁡(u)−1)tan2⁡(u)du{\displaystyle 2\int (\sec ^{2}(u)-1)\tan ^{2}(u)du} , Separate the integrand to make anti-differentiation easier. 2∫tan2⁡(u)sec2⁡(u)du−2∫tan2⁡(u)du{\displaystyle 2\int {\tan ^{2}(u)\sec ^{2}(u)}du-2\int {\tan ^{2}(u)}du} , Then dw=sec2⁡(u)du{\displaystyle dw=\sec ^{2}(u)du} Substitution yields 2∫w2dw+2∫tan2⁡(u)du{\displaystyle 2\int w^{2}dw+2\int \tan ^{2}(u)du} , ∫w2dw+∫sec2⁡(u)−1du{\displaystyle \int w^{2}dw+\int \sec ^{2}(u)-1du} , u=y{\displaystyle u={\sqrt {y}}} 23tan3⁡(y)−2tan⁡(y)−2y+C{\displaystyle {\frac {2}{3}}\tan ^{3}({\sqrt {y}})-2\tan({\sqrt {y}})-2{\sqrt {y}}+C} ,,,,, Let u=1θ{\displaystyle u={\frac {1}{\theta }}}.
   
   Then −du=1θ2{\displaystyle
   -du={\frac {1}{\theta ^{2}}}}.
   
   Substitution yields −∫sec4⁡(u)tan3⁡(u)du{\displaystyle
   -\int \sec ^{4}(u)\tan ^{3}(u)du} , Rearrange such that we can replace a term with a Pythagorean Identity.
   
   Note, sec⁡(u)tan⁡(u){\displaystyle \sec(u)\tan(u)} is the elementary derivative of sec⁡(u){\displaystyle \sec(u)}. −∫sec3⁡(u)tan2⁡(u)sec⁡(u)tan⁡(u)du{\displaystyle
   -\int \sec ^{3}(u)\tan ^{2}(u)\sec(u)\tan(u)du} ,, Keep sec⁡(u)tan⁡(u){\displaystyle \sec(u)\tan(u)} intact.
   
   Remember, we want to have a trig function and it's derivative in the integrand to employ a u-sub. ∫(sec5⁡(u)−sec3⁡(u))sec⁡(u)tan⁡(u)du{\displaystyle \int (\sec ^{5}(u)-\sec ^{3}(u))\sec(u)\tan(u)du} , Let w=sec⁡(u){\displaystyle w=\sec(u)}.
   
   Then dw=sec⁡(u)tan⁡(u)du{\displaystyle dw=\sec(u)\tan(u)du} Substitution yields −∫(w5−w3)dw{\displaystyle
   -\int (w^{5}-w^{3})dw} , −+C{\displaystyle
   -+C} , −+C{\displaystyle
   -+C} ,,,,, Let u=t2{\displaystyle u=t^{2}}.
   
   Then du2=tdt{\displaystyle {\frac {du}{2}}=tdt} Substitution yields 12∫sec4⁡(u)du{\displaystyle {\frac {1}{2}}\int \sec ^{4}(u)du} , Rearrange such that we can use a Pythagorean Identity. 12∫sec2⁡(u)sec2⁡(u)du{\displaystyle {\frac {1}{2}}\int \sec ^{2}(u)\sec ^{2}(u)du} , Keep one sec2⁡(u){\displaystyle \sec ^{2}(u)} intact.
   
   We will need for a u-sub. 12∫(1+tan2⁡(u))sec2⁡(u)du{\displaystyle {\frac {1}{2}}\int (1+\tan ^{2}(u))\sec ^{2}(u)du} , Let w=tan⁡(u){\displaystyle w=\tan(u)}.
   
   Then dw=sec2⁡(u){\displaystyle dw=\sec ^{2}(u)} Substitution yields 12∫1+w2dw{\displaystyle {\frac {1}{2}}\int 1+w^{2}dw} , 12+C{\displaystyle {\frac {1}{2}}+C} , w=tan⁡(u){\displaystyle w=\tan(u)} and u=t2{\displaystyle u=t^{2}} tan⁡(t2)2+tan⁡(t2)6+C{\displaystyle {\frac {\tan(t^{2})}{2}}+{\frac {\tan(t^{2})}{6}}+C}
3. 3

   Step 3: Rewrite all the rest in terms of cos⁡(x){\displaystyle \cos(x)} and/or sin⁡(x){\displaystyle \sin(x)} using Pythagorean Identities.

4. 4

   Step 4: Use algebra

5. 5

   Step 5: substitution

6. 6

   Step 6: and trig identities when applicable.

7. 7

   Step 7: Evaluate ∫sin5⁡(lnx)xdx{\displaystyle \int {\frac {\sin ^{5}(lnx)}{x}}dx}.

8. 8

   Step 8: Do a u-sub.

9. 9

   Step 9: Manipulate the integrand algebraically.

10. 10

    Step 10: Use a Pythagorean Identity to get sin2⁡(u){\displaystyle \sin ^{2}(u)} in terms of cosine.

11. 11

    Step 11: Do a u-sub.

12. 12

    Step 12: Manipulate the integrand algebraically.

13. 13

    Step 13: Integrate.

14. 14

    Step 14: Re-sub.

15. 15

    Step 15: Use a Double-Angle Identity for sin2⁡(x){\displaystyle \sin ^{2}(x)} and/or cos2⁡(x){\displaystyle \cos ^{2}(x)}

16. 16

    Step 16: Use algebra

17. 17

    Step 17: substitution

18. 18

    Step 18: and trig identities when applicable.

19. 19

    Step 19: Evaluate ∫sin2⁡(7x)cos2⁡(7x)dx{\displaystyle \int \sin ^{2}(7x)\cos ^{2}(7x)dx}.

20. 20

    Step 20: Do a u-sub.

21. 21

    Step 21: Use a Double-Angle Identity for sin2⁡(u){\displaystyle \sin ^{2}(u)} and cos2⁡(u){\displaystyle \cos ^{2}(u)}.

22. 22

    Step 22: Manipulate the integrand algebraically.

23. 23

    Step 23: Use a Double-Angle Identity for cos2⁡(2u){\displaystyle \cos ^{2}(2u)}.

24. 24

    Step 24: Manipulate the integrand algebraically.

25. 25

    Step 25: Integrate and re-sub.

26. 26

    Step 26: Separate a sin2⁡(x){\displaystyle \sin ^{2}(x)}.

27. 27

    Step 27: Rewrite all the rest in terms of cos⁡(x){\displaystyle \cos(x)} and/or sin⁡(x){\displaystyle \sin(x)} using Pythagorean Identities.

28. 28

    Step 28: Use algebra

29. 29

    Step 29: substitution

30. 30

    Step 30: and trig identities when applicable.

31. 31

    Step 31: Evaluate ∫πcos4x2dx{\displaystyle \int _{0}^{\pi }\cos ^{4}{\frac {x}{2}}dx}.

32. 32

    Step 32: Do a u-sub.

33. 33

    Step 33: Manipulate the integrand algebraically.

34. 34

    Step 34: Use a Double-Angle Identity for cos2⁡(u){\displaystyle \cos ^{2}(u)} 2∫0π22du{\displaystyle 2\int _{0}^{\frac {\pi }{2}}{}^{2}du}

35. 35

    Step 35: Manipulate the integrand algebraically.

36. 36

    Step 36: Use a Double-Angle Identity for cos2⁡(2u){\displaystyle \cos ^{2}(2u)}.

37. 37

    Step 37: Manipulate the integrand algebraically.

38. 38

    Step 38: Integrate.

39. 39

    Step 39: Compute the integral.

40. 40

    Step 40: Separate a tan⁡(x){\displaystyle \tan(x)}.

41. 41

    Step 41: Use Pythagorean Identities to get all else in terms of secant.

42. 42

    Step 42: Use algebra

43. 43

    Step 43: substitution

44. 44

    Step 44: and trig identities when applicable.

45. 45

    Step 45: Evaluate ∫tan4yydy{\displaystyle \int {\frac {\tan ^{4}{\sqrt {y}}}{\sqrt {y}}}dy}.

46. 46

    Step 46: Do a u-sub.

47. 47

    Step 47: Manipulate the integrand algebraically.

48. 48

    Step 48: Use a Pythagorean Identity for tan2⁡(x){\displaystyle \tan ^{2}(x)}.

49. 49

    Step 49: Manipulate the integrand algebraically.

50. 50

    Step 50: Do a u-sub on the first integral Let w=tan⁡(u){\displaystyle w=\tan(u)}.

51. 51

    Step 51: Use a Pythagorean Identity on the second integral.

52. 52

    Step 52: Integrate and re-sub.

53. 53

    Step 53: Separate a tan⁡(x){\displaystyle \tan(x)}.

54. 54

    Step 54: Use Pythagorean Identities to get all else in terms of secant.

55. 55

    Step 55: Substitute a sec⁡(x){\displaystyle \sec(x)}.

56. 56

    Step 56: Evaluate ∫sec4⁡(1θ)tan3⁡(1θ)1θ2dθ{\displaystyle \int \sec ^{4}({\frac {1}{\theta }})\tan ^{3}({\frac {1}{\theta }}){\frac {1}{\theta ^{2}}}d\theta }

57. 57

    Step 57: Do a u-sub.

58. 58

    Step 58: Manipulate the integrand algebraically.

59. 59

    Step 59: Use a Pythagorean Identity for tan2⁡(u){\displaystyle \tan ^{2}(u)} ∫sec3⁡(u)(sec2⁡(u)−1)sec⁡(u)tan⁡(u)du{\displaystyle \int \sec ^{3}(u)(\sec ^{2}(u)-1)\sec(u)\tan(u)du}

60. 60

    Step 60: Manipulate the integrand algebraically.

61. 61

    Step 61: Do a u-sub.

62. 62

    Step 62: Integrate.

63. 63

    Step 63: Re-sub.

64. 64

    Step 64: Separate a sec2⁡(x){\displaystyle \sec ^{2}(x)}.

65. 65

    Step 65: Use a Pythagorean Identity to get all else in terms of tangent.

66. 66

    Step 66: Substitute a tan⁡(x){\displaystyle \tan(x)} when applicable.

67. 67

    Step 67: Evaluate ∫tsec4⁡(t2)dt{\displaystyle \int t\sec ^{4}(t^{2})dt}

68. 68

    Step 68: Do a u-sub.

69. 69

    Step 69: Manipulate the integrand algebraically.

70. 70

    Step 70: Use a Pythagorean Identity for sec2⁡(u){\displaystyle \sec ^{2}(u)}.

71. 71

    Step 71: Do a u-sub.

72. 72

    Step 72: Integrate.

73. 73

    Step 73: Re-sub.

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