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Applied Calculus ILecture 29
Integrals of trigonometric functionsWe shall continue learning substitutions by considering integrals involvingtrigonometric functions.
Integrals of trigonometric functionsWe shall continue learning substitutions by considering integrals involvingtrigonometric functions.
Find
∫sin (x2 − 3)xdx
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.
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
Examples
Find
∫tanxdx
Examples
Find
∫tanxdx
Notice that
∫tanxdx =
∫sinx
cosxdx. The problematic part is the
denominator. So, let u = cosx. Then du = − sinxdx.
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
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
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.
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
ExamplesFind
∫dx
x lnx
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.
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).
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
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
Examples
Find
∫x83x
2+1dx
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
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
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
ExamplesFind
∫cos3 x
sinx+ 1dx
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.
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.
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
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.
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.
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.
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
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.
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