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Page 1: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

Applied Calculus ILecture 29

Page 2: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

Integrals of trigonometric functionsWe shall continue learning substitutions by considering integrals involvingtrigonometric functions.

Page 3: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

Integrals of trigonometric functionsWe shall continue learning substitutions by considering integrals involvingtrigonometric functions.

Find

∫sin (x2 − 3)xdx

Page 4: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 5: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 6: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

Examples

Find

∫tanxdx

Page 7: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

Examples

Find

∫tanxdx

Notice that

∫tanxdx =

∫sinx

cosxdx. The problematic part is the

denominator. So, let u = cosx. Then du = − sinxdx.

Page 8: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 9: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 10: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 11: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 12: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

ExamplesFind

∫dx

x lnx

Page 13: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 14: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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).

Page 15: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 16: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 17: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

Examples

Find

∫x83x

2+1dx

Page 18: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 19: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 20: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 21: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

ExamplesFind

∫cos3 x

sinx+ 1dx

Page 22: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 23: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 24: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 25: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 26: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 27: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 28: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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

Page 29: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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.

Page 30: Applied Calculus I - math.hawaii.edumath.hawaii.edu/~yury/spring2016/math215/rc/lec29.pdfApplied Calculus I Lecture 29. Integrals of trigonometric functions We shall continue learning

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