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Integral Rule
(Because)(1)sinxdx=−cosx+C
(1’)sinkxdx=−+C
(1”)sin(ax+b)dx=−+C
(2)cosxdx=sinx+C
(2’)coskxdx=+C
(2”)cos(ax+b)dx=+C
(3)tanxdx=−log|cosx|+C
(3’)tankxdx=−+C
(3”)tan(ax+b)dx=−+C
(1)← (cosx)=−sinx→(−cosx)=sinx sinxdx=−cosx+C (1’)← According to the Derivatives of Composite Functions = Let y=cosu, u=kx =−sinu×k=−sinkx×k (coskx)=−sinkx×k (−)=sinkx Hence sinkxdx=−+C Similarly, (1”) will be obtained.
Note that (ax+b)=a (constant term vanishes).
(2)←So the denominator of the fraction − never contains the constant term b. (sinx)=cosx cosxdx=sinx+C (2’)← Let y=sinu, u=kx =cosu×k=coskx×k (sinkx)=coskx×k ()=coskx Hence coskxdx=+C Similarly, (2”) will be obtained.
Note that the denominator never contains the constant term b.
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(4)dx=−+C=−cotx+C
(4’)dx=−+C=−+C
(4”)dx=−+C
(5)dx=tanx+C=−+C
(5’)dx=+C
(5”)dx=+C
(3)←
log|x|=
(cosx)=−sinx
So, let y=log|u|, u=cosx
==×(−sinx)=×(−sinx)=−tanx
(log|cosx|)=−tanx
(−log|cosx|)=tanx
tanxdx=−log|cosx|+C
(3’)←Let y=log|s|, s=cost, t=kx ==(−sint)×k=×(−sint)×k (log|coskx|)=−tankx×k (−)=tankx Hence tankxdx=−+C Similarly, (3”) will be obtained.
Note that the denominator never contains the constant term b.
(4)←
=cotx=
( )=
So
(cotx)==−
(−cotx)=
Hencedx=−cotx+C (4’)(4”) Omitted. (5)←
tanx=
( )=
So
(tanx)==
Hencedx=tanx+C (5’)(5”) Omitted. |
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Examples (1’)sin2xdx=−+C (1”)sin(2x+1)dx=−+C (2’)cos2xdx=+C (2”)cos(2x+1)dx=+C |
(3’)tan2xdx=−+C (4’)dx=−+C (5’)dx=+C |
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Using the following rules the trigonometric functions including multiplications/squares can be converted into the additions/subtractions of the functions.
(A)sinαcosβ={sin(α+β)+sin(α−β)}
(B)cosαsinβ={sin(α+β)−sin(α−β)} (C)cosαcosβ={cos(α+β)+cos(α−β)} (D)sinαsinβ=−{cos(α+β)−cos(α−β)} (E)sin2α= (F)cos2α= (G)1+tan2α=→tan2α=−1→(5) (H)1+=→=−1→(4) |
Examples (A)sin4xcos2xdx={sin6x+sin2x}dx ←(1’) ={−−}+C=−−+C (C)cos4xcos2xdx={cos6x+cos2x}dx ←(2’) ={+}+C=++C (D)sin4xsin2xdx=−{cos6x−cos2x}dx ←(2’) =−{−}+C=−++C (E)sin22xdx=dx=(x−)+C =x−+C (F)cos22xdx=dx=(x+)+C =x++C (G)tan22xdx=(−1)dx=−x+C
It looks starnge that the degree(power) of the integral is less than its integrand(function to be integrated) concerning tan2x, but it’s true.
(H)dx=(−1)dx=−−x+C |
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Question 1Find the following integrals. (Choose the correct answer. Maybe these are difficult only at a glance, you may use your own calculation sheet.) ++C −+C +C −+C +C −+C |
(1’)sinkxdx=−+C
→sin3xdx=−+C
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++C −+C +C −+C +C −+C |
(E)sin2α=
→sin23x=sin23xdx=()dx ←(2’) =(x−)+C=−+C |
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++C −+C +C −+C +C −+C |
(4”)dx=−+C
→dx=−+C |
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++C −+C +C −+C +C +C |
(2’)coskxdx=+C
→cos3xdx=+C |
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++C −+C +C −+C +C +C |
(F)cos2α=
→cos23x=cos23xdx=dx ←(2’) =(x+)+C=++C |
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++C −+C +C −+C +C +C |
(5”)dx=+C
→dx=+C |
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+x+C −x+C +x+C −−x+C +C −+C |
(3’)tankxdx=−+C
→tan3xdx=−+C |
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+x+C −x+C +x+C −−x+C +C −+C |
(G)1+tan2α=→tan2α=−1
→tan23xdx=(−1)dx ←(5’)=−x+C |
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+x+C −x+C +x+C −−x+C +C −+C |
(log|sin3x|)=×cos3x×3=
→()=
→dx=+C
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+x+C −x+C +x+C −−x+C +C −+C |
(H)1+=→=−1
→dx=(−1)dx ←(4’) =−−x+C |
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++C −+C −++C ++C −+C −−+C |
(D)sinαsinβ=−{cos(α+β)−cos(α−β)}
→sin5xsinx=−{cos6x−cos4x} sin5xsinx dx=−{cos6x−cos4x}dx ←(2’) =−(−)+C=−++C |
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++C −+C −++C ++C −+C −−+C |
(A)sinαcosβ={sin(α+β)+sin(α−β)}
→sin5xcosx={sin6x+sin4x} sin5xcosx dx={sin6x+sin4x}dx ←(1’) =(−−)+C=−−+C |
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++C −+C −++C ++C −+C −−+C |
(C)cosαcosβ={cos(α+β)+cos(α−β)}
→cos5xcosx={cos6x+cos4x} cos5xcosx dx={cos6x+cos4x}dx ←(2’) =(+)+C=++C |