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

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