Question

Find the Indefinite Integral...

Answer 1

\[\int\limits_{}^{}\sec(1-x)\tan(1-x)dx\]

Answer 2

\(\color{#000000 }{ \displaystyle \frac{ d}{dx} \sec(x)=? }\)

Answer 3

secxtanx

Answer 4

Yes, very good

Answer 5

you know that integrating is a reverse operation of differentiation. Right?

Answer 6

I believe so

Answer 7

So, since

\(\color{#000000 }{ \displaystyle \frac{ d}{dx} \sec(x)=\sec(x)\tan(x) }\)

therefore,

\(\color{#000000 }{ \displaystyle \int \sec(x)\tan(x)~dx=\sec(x)\color{grey}{\rm +C} }\)

Answer 8

have you done any u-substitution before?

Answer 9

yes, so my u would be (1-x) and my du would be (-dx)

Answer 10

Yes, very good

u=1-x
du= - dx

Answer 11

But, since you don't have the negative inside the integral, you may rearrange the du=-dx.

You can divide both sides by -1, and you get,

-du=dx

Answer 12

So, you have
u = 1-x
-du = dx

and you can substitute this right into the problem.

Answer 13

\(\color{#000000 }{ \displaystyle \int \sec(1-x)\tan(1-x)~dx }\)

and with your substitution,
u = 1-x
-du = dx

the integral will become,

\(\color{#000000 }{ \displaystyle \int \sec(u)\tan(u)~(-du) }\)

and then this is same as,

\(\color{#000000 }{ \displaystyle -\int \sec(u)\tan(u)~du }\)

Answer 14

if something didn't make sense, or you have an additional question about this, then ask.

Also, don't forget to substitute the x back instead of u.
(u=1-x \(\Longrightarrow\) x=1-u)

Answer 15

ok that makes sense so far

Answer 16

Alright integrate (with respect to "u"), and then substitue the x back in

Answer 17

would my final answer be -sec(1-x)+c ?

Answer 18

Yes, very good!

Answer 19

Woo! would you mind walking me through another problem?

Answer 20

Sure, although I am also pretty sure, that for the next problem you will need no more than a hint or two:)

Answer 21

\[\int\limits_{}^{}\frac{ sinx }{ \cos^3x}dx\]

Answer 22

\(\color{#000000 }{ \displaystyle \frac{\sin x}{\cos^3x}=(\cos x)^{-3}\times (\sin x) }\)

Answer 23

Seems as though that the integrand was obtained from differentiating some function with a chain rule?

do you see what I am taking about?

Answer 24

I think so

Answer 25

well, we can solve it through u-sub rather than using "recognizing derivatives" technique...

Answer 26

I don't know what your instructor told you to do, but which approach would be harder for you?

Answer 27

As reagrds to recognizing derivatives, (\(n\ne 0\))

\(\color{#000000 }{ \displaystyle \frac{d}{dx}(\cos x)^n=n(\cos x) ^{n-1} \times (-\sin x) =-n\cdot \cos^{n}x\sin x}\)

Answer 28

for negative n,

\(\color{#000000 }{ \displaystyle \frac{d}{dx}(\cos x)^{-n}=-n(\cos x) ^{-n-1} \times (-\sin x) =n\cdot \cos^{n}x\sin x}\)

Answer 29

We have mainly worked through u-sub

Answer 30

ok, so let's do the u sub

Answer 31

if you see something in a form of,

\(\color{#000000 }{ \displaystyle \int {\rm G}\left[f(x)\right]\times f'(x)~dx }\)

you would set,

\(\color{#000000 }{ \displaystyle u=f(x) }\)
\(\color{#000000 }{ \displaystyle du=f'(x)~dx }\)

and then you would have the following,
\(\color{#000000 }{ \displaystyle \int {\rm G}\left[u\right]~du }\)

Answer 32

Right? ... just the concept of you substitution when the problem is in that form.

Answer 33

In your case, you are dealing with the same thing,

\(\color{#000000 }{ \displaystyle \int \frac{1}{(\cos x)^3}\times \sin x ~dx }\)

just,. that you don't have the negative in front of sin(x), but we will resolve that the same way as we did in problem 1, when we said -du=dx.

Answer 34

I got my u to be cos x and my du to be -sinxdx -->
then I divided both sides by -1 -->
to get this \[-\int\limits_{}^{}u^{-3} du\]
then by substituting in I got -1/2sec^2+c

Answer 35

\(\color{#000000 }{ \displaystyle -\int u^{-3} du }\)

\(\color{#000000 }{ \displaystyle = \frac{-u^{-3+1} }{-3+1}=\frac{1}{2}u^{-2}=\frac{1}{2}\cos^{-2}x=\frac{1}{2}\sec^2x+C }\)

Answer 36

yes, fabulous !

Answer 37

thanks for the help man!

Answer 38

Anytime!
Good work!

Also, by the way, you can divide by any number when you solve the "du" part, whether this number is -1, like we had twice, or if this 2, e, pi, psi, or any other constant .... (even by variables)

Answer 39

time to finish these last couple of problems!

Answer 40

ok

Answer 41

Any questions?

Answer 42

let me look through my remaining homework problems real quick

Answer 43

Alright :)

Answer 44

I will just put up an example of what I mean by "dividing by variables" when solving the "du" part....

Answer 45

\[\int\limits_{}^{} x \sqrt{x+6}dx and \int\limits_{0}^{1}x \sqrt{1-x^2}dx\]

Answer 46

So, suppose you have:

\(\color{#000000 }{ \displaystyle \int e^\sqrt{x} dx }\)

then, I set,

\(\color{#000000 }{ \displaystyle u=\sqrt{x} }\) \(\tiny \\[0.8em]\)
\(\color{#000000 }{ \displaystyle du=\frac{1}{2\sqrt{x}}dx }\)

but, I don't have the \(\color{#000000 }{ \displaystyle \frac{1}{2\sqrt{x}} }\) inside my integral, so what do I do?

I will multiply time \(2\sqrt{x}\) on both sides of the du part, to get,

\(\color{#000000 }{ \displaystyle 2\sqrt{x} du=dx }\)

then, I know that \(u=\sqrt{x}\) (this is my substitution)
So, I can say,

\(\color{#000000 }{ \displaystyle 2u~du=dx }\)

So, the old problem that I had,

\(\color{#000000 }{ \displaystyle \int e^\sqrt{x} dx }\)


would turn into,

\(\color{#000000 }{ \displaystyle \int e^u (2u~du)\quad \Longrightarrow \quad 2\int u~e^u~du }\)

and that would be solved via "integration by parts" wqhich you might not have learned yet, but this is an example of playing with variables..... (and at leas this is integratable, unlike the initial integrand)

Answer 47

that was just an example of playing with variables in a u-substitution... there is a lot of things you can do there ... (but still, integrals are not manipulable like limits)

Answer 48

ok thanks for the advice!

Answer 49

\(\color{#000000 }{ \displaystyle \int x\sqrt{x+6}dx }\)

Answer 50

if you had something like
\(\color{#000000 }{ \displaystyle \int (z-a)\sqrt{z}~dz }\)

then, wouldn't tha be easier to integrate?
(expand and apply the power rule)

Answer 51

you might want to tr to reach something similar ...
what do you think your u-substitution should be?

Answer 52

my u was x+6, my du was dx, and my x was u-6 giving me -->
\[\int\limits_{}^{}(u-6)u^{1/2}du \]

Answer 53

Yes, that is briliant !

Answer 54

now, multiply it out, integrtae, and after that substitute the x back

Answer 55

distributing i got -->
\[\int\limits_{}^{}u^{3/2}-6u^{1/2 }du\]

Answer 56

final answer is -->
\[\frac{ 2 }{ 5 }(x+6)^{5/2}-4(x+6)^{3/2}=C\]

Answer 57

yes, that = was supposed t be + :D

Answer 58

my keyboard also has that sometimes :)

Answer 59

good job

Answer 60

no thank you! your explanations have been very helpful

Answer 61

your next problem, unless you have questions about something previous, is

\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1} x\sqrt{1-x^2}dx }\)

Answer 62

There are two ways to do u-substitution with definite integrals.

Answer 63

1.
You susbtitute just like you do by indefinite integrals,
THEN, you integrate and substitute the original variable back.
THEN, you evaluate at the given limits of integration.

Answer 64

2.
You substitute just like by indefinite integrals, \(\color{red}{\bf AND}\) you change the limits of integration.

So, if you have u=f(x), and limits of integration x=a and x=b, then your new limits would be f(a) and f(b). \(\color{red}{\bf BUT}\) after you integrate, you \(\color{red}{\bf don't}\) substitute the original variable back; \(\color{red}{\bf rather}\), you evaluate at the new limits of integration (at u=f(a), and u=f(b).)

Answer 65

So, you are used to doing it the first way I would expect, right?

Answer 66

yes

Answer 67

So let's do it way 1 together, and then I will put up the way 2 afterwords.
Suffices?

Answer 68

cool beans

Answer 69

Alright!

Answer 70

Recall that:

if you see an integral in a form of, \(\color{#000000 }{ \displaystyle \int {\rm G}\left[f(x)\right]\times f'(x)~dx }\)

you would set, \(\color{#000000 }{ \displaystyle u=f(x) }\)
\(\color{#000000 }{ \displaystyle du=f'(x)~dx }\)

and then you would have, \(\color{#000000 }{ \displaystyle \int {\rm G}\left[u\right]~du }\)

And recall too that we can multiplyand divide by numbers in the "du" part as we like.

Answer 71

Your problem is: \(\color{#000000 }{ \displaystyle \int\limits_{0}^{1} x\sqrt{1-x^2}dx }\)

I am sure you know the derivative of \(1-x^2\).

Answer 72

2xdx

Answer 73

so what is your u, and wat is your du, exactly ?

Answer 74

u is 1-x^2 and my du is 2xdx

Answer 75

du is a little off

Answer 76

what is the derivative of \(\color{red}{-}x^2\) ?

Answer 77

whoops my du would be -2xdx

Answer 78

yes \(\color{#000000 }{ \displaystyle u=1-x^2 }\)
\(\color{#000000 }{ \displaystyle du=-2x~dx }\)

Answer 79

and since it is negative, would i flip the range?

Answer 80

you don't have the "2" coefficient in the integral, so divide both sides of the "du" by 2.

Answer 81

\[\int\limits\limits_{1}^{0} \]

Answer 82

Well, if this is what you mean, then

\(\color{#000000}{\displaystyle\int\limits_{a}^{b} f(x)~dx=-\int\limits_{b}^{a} f(x)~dx}\)

but you don't need that

Answer 83

ok

Answer 84

\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1} x\sqrt{1-x^2}dx }\)

set, \(\color{#000000 }{ \displaystyle u=1-x^2 }\)
\(\color{#000000 }{ \displaystyle du=-2x~dx }\)

remember, we can play with "du", and we need to do this, since we don't have the -2 in the integral (right?)

So our du part will become, \(\color{#000000 }{ \displaystyle \frac{1}{-2}du=x~dx }\)

Answer 85

so just substitute this accordingly...

Answer 86

\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1} \color{#ff0000}{x}\sqrt{\color{#0000ff}{1-x^2}}\color{#ff0000}{dx} }\)

set, \(\color{#0000ff }{ \displaystyle u=1-x^2 }\)
\(\color{#ff0000 }{ \displaystyle du=-2x~dx }\)

remember, we can play with "du", and we need to do this, since we don't have the -2 in the integral (right?)
So our du part will become, \(\color{#ff0000 }{ \displaystyle \frac{1}{-2}du=x~dx }\)


And your integral will be,

\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1} \sqrt{\color{#0000ff}{u~}}\left(\color{#ff0000}{\frac{1}{-2}du}\right) }\)

which simplifies to,

\(\color{#000000 }{ \displaystyle -\frac{1}{2}\int\limits_{0}^{1} \sqrt{u~}du }\)

Answer 87

ok i understand so far

Answer 88

and from there, I am pretty sure you can integrate it yourself (power rule)

Answer 89

but, when you find the integral, first substitute the x back into the equation, and then evaluate at the given limits of integration.

Answer 90

if you need help with anything let me know.

Answer 91

is my final answer 1/3?

Answer 92

yes, very good!

Answer 93

Want to see way 2 f you-substitution?

Answer 94

sure, it wouldn't hurt

Answer 95

\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1} x\sqrt{1-x^2}dx }\)


you set, \(\color{#000000 }{ \displaystyle u=1-x^2 }\)
\(\color{#000000 }{ \displaystyle du=-2x~dx\quad \Longrightarrow \quad \frac{1}{-2}du=x~dx }\)

And the limits of integration, change as follows:
\(\color{#000000 }{ \displaystyle x=\color{blue}{0}\quad \quad \Rightarrow\quad\quad u=1-\color{blue}{0}^2 \quad \quad \Rightarrow\quad\quad u=1 }\)
\(\color{#000000 }{ \displaystyle x=\color{blue}{1}\quad \quad \Rightarrow\quad\quad u=1-\color{blue}{1}^2 \quad \quad \Rightarrow\quad\quad u=0 }\)
\(\tiny\\[0.5em]\)

OLD INTEGRAL \(\color{#000000 }{ \displaystyle \Longrightarrow \Longrightarrow }\) NEW INTEGRAL\(\tiny\\[1.0em]\)
\(\color{#000000 }{ \displaystyle \int\limits_{0}^{1} x\sqrt{1-x^2}dx }\) \(\color{#000000 }{ \displaystyle \Longrightarrow \Longrightarrow }\) \(\color{#000000 }{ \displaystyle \frac{1}{-2}\int\limits_{1}^{0} \sqrt{~u~}du }\)

I got the limits 1 and 0, NOT BECAUSE I FLIPPED THEM!
(THIS IS IMPORTANT NOT TO MISS, ISN'T IT? :D)

And then I can integrate this easily,

\(\color{#000000 }{ \displaystyle \frac{1}{-2}\int\limits_{1}^{0} \sqrt{~u~}du }\)

\(\color{#000000 }{ \displaystyle -\frac{1}{2}\int\limits_{1}^{0} \sqrt{~u~}du }\)

let's use that propery with flipping the limits :)

\(\color{#000000 }{ \displaystyle \frac{1}{2}\int\limits_{0}^{1} \sqrt{~u~}du }\)

\(\color{#000000 }{ \displaystyle \frac{1}{2}\int\limits_{0}^{1} u^{1/2} du =\frac{1}{2}\left( \left. \frac{2}{3} u^{3/2}\right|^{u=1}_{u=0} \right) =\frac{1}{2}\left(\frac{2}{3} (1)^{3/2}-\frac{2}{3} (0)^{3/2}\right)}\)

\(\color{#000000 }{ \displaystyle =\frac{1}{2}\left(\frac{2}{3} -0\right)=\frac{1}{2}\times \frac{2}{3}=\frac{1}{3}}\)

Answer 96

if you need more material on this stuff let me know.

Answer 97

screen shotted this stuff for later!

Answer 98

Also, there is an alternative way of doing this integral, and that is, via "trig substitution"