Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied...
Transcript of Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied...
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Applied Calculus ILecture 29
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Integrals of trigonometric functionsWe shall continue learning substitutions by considering integrals involvingtrigonometric functions.
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Integrals of trigonometric functionsWe shall continue learning substitutions by considering integrals involvingtrigonometric functions.
Find
∫sin (x2 − 3)xdx
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Integrals of trigonometric functionsWe shall continue learning substitutions by considering integrals involvingtrigonometric functions.
Find
∫sin (x2 − 3)xdx
The expression that seems to complicate things here is x2 − 3. So, let usmake the substitution u = x2 − 3. Then du = 2xdx, whic impliesxdx = 1
2du.
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Integrals of trigonometric functionsWe shall continue learning substitutions by considering integrals involvingtrigonometric functions.
Find
∫sin (x2 − 3)xdx
The expression that seems to complicate things here is x2 − 3. So, let usmake the substitution u = x2 − 3. Then du = 2xdx, whic impliesxdx = 1
2du.
Then our integral becomes∫1
2sinudu = −1
2cosu+ C = −1
2cos (x2 − 3) + C
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Examples
Find
∫tanxdx
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Examples
Find
∫tanxdx
Notice that
∫tanxdx =
∫sinx
cosxdx. The problematic part is the
denominator. So, let u = cosx. Then du = − sinxdx.
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Examples
Find
∫tanxdx
Notice that
∫tanxdx =
∫sinx
cosxdx. The problematic part is the
denominator. So, let u = cosx. Then du = − sinxdx.
Now we see that our integral becomes∫−duu
= − ln |u|+ C = − ln | cosx|+ C
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Examples
Find
∫tanxdx
Notice that
∫tanxdx =
∫sinx
cosxdx. The problematic part is the
denominator. So, let u = cosx. Then du = − sinxdx.
Now we see that our integral becomes∫−duu
= − ln |u|+ C = − ln | cosx|+ C
Find
∫cotxdx
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Examples
Find
∫tanxdx
Notice that
∫tanxdx =
∫sinx
cosxdx. The problematic part is the
denominator. So, let u = cosx. Then du = − sinxdx.
Now we see that our integral becomes∫−duu
= − ln |u|+ C = − ln | cosx|+ C
Find
∫cotxdx
Notice that
∫cotxdx =
∫cosx
sinxdx. The problematic part is the
denominator. So, let u = sinx. Then du = cosxdx.
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Examples
Find
∫tanxdx
Notice that
∫tanxdx =
∫sinx
cosxdx. The problematic part is the
denominator. So, let u = cosx. Then du = − sinxdx.
Now we see that our integral becomes∫−duu
= − ln |u|+ C = − ln | cosx|+ C
Find
∫cotxdx
Notice that
∫cotxdx =
∫cosx
sinxdx. The problematic part is the
denominator. So, let u = sinx. Then du = cosxdx.
Now we see that our integral becomes∫du
u= ln |u|+ C = ln | sinx|+ C
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ExamplesFind
∫dx
x lnx
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ExamplesFind
∫dx
x lnx
Again, the part that complicates things is the denominator. But if we trythe substitution u = x lnx we see that du = (lnx+ 1)dx, and then wecan’t express everything in terms of u.
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ExamplesFind
∫dx
x lnx
Again, the part that complicates things is the denominator. But if we trythe substitution u = x lnx we see that du = (lnx+ 1)dx, and then wecan’t express everything in terms of u.
The idea is to make a substitution for the part of your integrand whosederivative gives you the rest of the integrand (perhaps multiplied by aconstant).
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ExamplesFind
∫dx
x lnx
Again, the part that complicates things is the denominator. But if we trythe substitution u = x lnx we see that du = (lnx+ 1)dx, and then wecan’t express everything in terms of u.
The idea is to make a substitution for the part of your integrand whosederivative gives you the rest of the integrand (perhaps multiplied by aconstant).
In this case, we notice that (lnx)′ = 1x
. So, let us try u = lnx. Thendu = 1
xdx, and our integral becomes
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ExamplesFind
∫dx
x lnx
Again, the part that complicates things is the denominator. But if we trythe substitution u = x lnx we see that du = (lnx+ 1)dx, and then wecan’t express everything in terms of u.
The idea is to make a substitution for the part of your integrand whosederivative gives you the rest of the integrand (perhaps multiplied by aconstant).
In this case, we notice that (lnx)′ = 1x
. So, let us try u = lnx. Thendu = 1
xdx, and our integral becomes∫
du
u= ln |u|+ C = ln | lnx|+ C
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Examples
Find
∫x83x
2+1dx
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Examples
Find
∫x83x
2+1dx
Let u = 3x2 + 1, then du = 6xdx. Hence,∫x83x
2+1dx =1
6
∫8udu =
1
6· 8
u
ln 8+ C =
83x2+1
6 ln 8+ C
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Examples
Find
∫x83x
2+1dx
Let u = 3x2 + 1, then du = 6xdx. Hence,∫x83x
2+1dx =1
6
∫8udu =
1
6· 8
u
ln 8+ C =
83x2+1
6 ln 8+ C
Find
∫sin7 x cosxdx
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Examples
Find
∫x83x
2+1dx
Let u = 3x2 + 1, then du = 6xdx. Hence,∫x83x
2+1dx =1
6
∫8udu =
1
6· 8
u
ln 8+ C =
83x2+1
6 ln 8+ C
Find
∫sin7 x cosxdx
Let u = sinx, then du = cosxdx, yielding∫sin7 x cosxdx =
∫u7du =
1
8u8 + C =
1
8sin8 x+ C
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ExamplesFind
∫cos3 x
sinx+ 1dx
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ExamplesFind
∫cos3 x
sinx+ 1dx
Proceeding as usual, let u = sinx+ 1, then du = cosxdx. So, cosxdxbecomes du, the denominator becomes u, but we still have the factorcos2 x that we need to express in terms of u.
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ExamplesFind
∫cos3 x
sinx+ 1dx
Proceeding as usual, let u = sinx+ 1, then du = cosxdx. So, cosxdxbecomes du, the denominator becomes u, but we still have the factorcos2 x that we need to express in terms of u.
Recall that cos2 x+ sin2 x = 1, so cos2 x = 1− sin2 x. Also,sin2 x = (sinx+ 1− 1)2 = (u− 1)2 = u2 − 2u+ 1. Therefore,cos2 x = 2u− u2.
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ExamplesFind
∫cos3 x
sinx+ 1dx
Proceeding as usual, let u = sinx+ 1, then du = cosxdx. So, cosxdxbecomes du, the denominator becomes u, but we still have the factorcos2 x that we need to express in terms of u.
Recall that cos2 x+ sin2 x = 1, so cos2 x = 1− sin2 x. Also,sin2 x = (sinx+ 1− 1)2 = (u− 1)2 = u2 − 2u+ 1. Therefore,cos2 x = 2u− u2.Thus, we obtain∫
cos3 x
sinx+ 1dx =
∫2u− u2
udu =
∫(2− u)du = 2u− 1
2u2 + C =
= 2(sinx+ 1)− 1
2(sinx+ 1)2 + C
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ExamplesFind
∫cos3 x
sinx+ 1dx
Proceeding as usual, let u = sinx+ 1, then du = cosxdx. So, cosxdxbecomes du, the denominator becomes u, but we still have the factorcos2 x that we need to express in terms of u.
Recall that cos2 x+ sin2 x = 1, so cos2 x = 1− sin2 x. Also,sin2 x = (sinx+ 1− 1)2 = (u− 1)2 = u2 − 2u+ 1. Therefore,cos2 x = 2u− u2.Thus, we obtain∫
cos3 x
sinx+ 1dx =
∫2u− u2
udu =
∫(2− u)du = 2u− 1
2u2 + C =
= 2(sinx+ 1)− 1
2(sinx+ 1)2 + C
In the hindsight, we could do∫cos3 x
sinx+ 1dx =
∫(1− sin2 x) cosx
1 + sin xdx =
∫(1− sinx) cosxdx
and then integrate.
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ExampleAn epidemic is growing in a region according to the rate
N ′(t) =100t
t2 + 2,
where N(t) is the number of people infected after t days. Find a formulafor the number of people infected after t dyas, given that 37 people wereinfected at t = 0.
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ExampleAn epidemic is growing in a region according to the rate
N ′(t) =100t
t2 + 2,
where N(t) is the number of people infected after t days. Find a formulafor the number of people infected after t dyas, given that 37 people wereinfected at t = 0.So, we need a formula for N(t) given that N(0) = 37. Clearly, N(t) is
some antiderivative of100t
t2 + 2. So, let’s find the whole family.
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ExampleAn epidemic is growing in a region according to the rate
N ′(t) =100t
t2 + 2,
where N(t) is the number of people infected after t days. Find a formulafor the number of people infected after t dyas, given that 37 people wereinfected at t = 0.So, we need a formula for N(t) given that N(0) = 37. Clearly, N(t) is
some antiderivative of100t
t2 + 2. So, let’s find the whole family.∫
100t
t2 + 2dt
u=t2+2,du=2tdt︷︸︸︷=
∫50
udu = 50 ln |u|+ C = 50 ln |t2 + 2|+ C
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ExampleAn epidemic is growing in a region according to the rate
N ′(t) =100t
t2 + 2,
where N(t) is the number of people infected after t days. Find a formulafor the number of people infected after t dyas, given that 37 people wereinfected at t = 0.So, we need a formula for N(t) given that N(0) = 37. Clearly, N(t) is
some antiderivative of100t
t2 + 2. So, let’s find the whole family.∫
100t
t2 + 2dt
u=t2+2,du=2tdt︷︸︸︷=
∫50
udu = 50 ln |u|+ C = 50 ln |t2 + 2|+ C
So, N(t) = 50 ln (t2 + 2) + C for some specific value of C. To find thisvalue, we need to use the fact that N(0) = 37.
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ExampleAn epidemic is growing in a region according to the rate
N ′(t) =100t
t2 + 2,
where N(t) is the number of people infected after t days. Find a formulafor the number of people infected after t dyas, given that 37 people wereinfected at t = 0.So, we need a formula for N(t) given that N(0) = 37. Clearly, N(t) is
some antiderivative of100t
t2 + 2. So, let’s find the whole family.∫
100t
t2 + 2dt
u=t2+2,du=2tdt︷︸︸︷=
∫50
udu = 50 ln |u|+ C = 50 ln |t2 + 2|+ C
So, N(t) = 50 ln (t2 + 2) + C for some specific value of C. To find thisvalue, we need to use the fact that N(0) = 37.
Plugging t = 0 into the formula for N(t) we get 50 ln 2 +C = 37 that is,C = 37− 50 ln 2. Therefore, N(t) = 50 ln (t2 + 2) + 37− 50 ln 2.
This can be simplified to N(t) = 50 ln(
t2
2+ 1)+ 37