Question

Use substitution to evaluate the integrals.
Problem below.

Answer 1

\[1.) \int\limits_{1}^{4} \frac{ dy }{ 2 \sqrt{y} (1+ \sqrt{y})^2 } \]
\[2.) \int\limits_{-1}^{-\frac { 1 }{ 2 }} t^{-2} \sin^2(1 +\frac{ 1 }{ t }) dt\]

Answer 2

@zepdrix

Answer 3

\[\large \int\limits\limits_{1}^{4} \frac{1}{(1+ \color{orangered}{\sqrt{y}})^2 }\left(\color{green}{\frac{1}{2\sqrt y}dy}\right)\]

Make sure you're familiar with this derivative, it will help quite a bit with identifying u-sub's like this one.\[\large (\sqrt x)'=\frac{1}{2\sqrt x}\]

Answer 4

All I did so far was a rearrange things a bit so It's a little easier to read.
It looks like in this problem we'll want the orange term to be our `u`.
Do you see where our du will show up?

Answer 5

so \[u=\sqrt{y}\]

Answer 6

mhmmm

Answer 7

\[\large du=?\]

Answer 8

\[du=\frac{ 1 }{ 2 \sqrt{x} }\]

Answer 9

That should be a y right? :o

Answer 10

oh yeah.. sorry. haha
\[du= \frac{ 1 }{ 2 \sqrt{y} }\]

Answer 11

Hmmmm interesting :O that looks like our green thingy-ma-bobber...

Answer 12

woops your du should be producing a dy somewhere:\[\large du= \frac{ 1 }{ 2 \sqrt{y} }dy\]

Answer 13

Ain't that a good thing? lol

Answer 14

so it's in terms of dy?

Answer 15

Yes that's a good thing.
That will allow us to change our integral like this:\[\large \int\limits\limits\limits_{y=1}^{4} \frac{1}{(1+ \color{orangered}{\sqrt{y}})^2 }\left(\color{green}{\frac{1}{2\sqrt y}dy}\right) \qquad\to\qquad \int\limits\limits\limits_{y=1}^{4} \frac{1}{(1+ \color{orangered}{u})^2 }\left(\color{green}{du}\right)\]

Answer 16

why is the integral at y-1 to 4? where did the y-1 come from?

Answer 17

y=1?

Answer 18

yeah why is it y=1? or it doesnt matter?

Answer 19

I put the y= to remind you that those boundaries only apply to y.

We either have to change our limits of integration to u=,
or we can simply undo our substitution after we integrate.

We just have to make sure we DO NOT plug in those y values for our u.

Answer 20

ohhhh, I see. okay. haha

Answer 21

Do you understand this step? How to integrate this function.
\[\large \int\limits_{y=1}^4 (1+u)^{-2}du\]
Just apply the power rule for integrals! :D

Answer 22

how did you get that? antiderive or derive?

Answer 23

How did I get what? +_+

Answer 24

(1+u)^-2 haha

Answer 25

Rules of Exponents:\[\large \frac{1}{(1+u)^2} \qquad=\qquad (1+u)^{-2}\]

Answer 26

ohhh -_- lol
where do we go from there?

Answer 27

do we plug in the subs?

Answer 28

take the anti-derivative D:

Answer 29

\[\frac{ 1+u ^{-2+1} }{ -2+1 } du\]

Answer 30

The -2 is an exponent on the entire term.
Not just on the u. That's why there are brackets :o

Answer 31

If you use brackets, then it looks ok.
\[\large \frac{(1+u)^{-2+1}}{-2+1}\]

Answer 32

Oh okay. so...
\[\frac{ (1+u)^{-1} }{ -1 } = -(1+u)^{-1}\]

Answer 33

Ok looks good.
From here we can undo our substitution, u=sqrt(y).\[\large -(1+\sqrt y)^{-1}\;|_1^4\]From here we can evaluate the function at our limits.

Answer 34

I forgot du at the end.... does that change the function?

Answer 35

\[\int\limits_{1}^{4} -(1+u)^{-1} du\]

Answer 36

The du disappears as a part of the anti-differentiation process, along with the squiggly bar.

Answer 37

oh that's right! hah

Answer 38

so..
\[-(1+ \sqrt{1})^{-1} - (-(1+ \sqrt{4})^{-1} = -2-(-1.5) = 0.5\]

Answer 39

Hmm it looks like 4 was our `upper limit of integration`.
We should be subtracting the sqrt(1) from that.

Answer 40

damn! I did it backwards. -.- lol

Answer 41

And also, don't forget about your -1 exponents!! D:

Answer 42

\[=> (-(1+ \sqrt{4})^{-1})-(-(1+ \sqrt{1})^{-1}) = (-\frac{ 1 }{ 3 })-(-\frac{ 1 }{ 2 }) = \frac{ 1 }{ 6 }\]

Answer 43

yay good job \c:/

Answer 44

sweet!(: how do we go about #2

Answer 45

let \[u=(1+\frac{ 1 }{ t })\]
@zep

Answer 46

\[\large \int\limits\limits_{-1}^{-\frac { 1 }{ 2 }} t^{-2} \sin^2\left(\color{orangered}{1+\frac{1}{t}}\right) dt\]

Ah yes good :o that's what we want for our u.

Answer 47

\[du=\frac{ -1 }{ t^2 }\]

Answer 48

you gotta remember your differential D:::

Answer 49

\[\large \frac{du}{dt}=-\frac{1}{t^2} \qquad\to\qquad du=-\frac{1}{t^2}dt\]

Answer 50

oh right! dt!

Answer 51

Multiply both sides by -1 to move it to the other side, giving us,\[\large -du=\frac{1}{t^2}dt \qquad\to\qquad -du=t^{-2}dt\]

Answer 52

so now....
\[\int\limits_{-1}^{-\frac{ 1 }{ 2 }} t^{-2} \sin^2(u) dt\]

Answer 53

\[\large \int\limits\limits_{-1}^{-\frac{ 1 }{ 2 }} t^{-2} \sin^2(u) dt \qquad=\qquad\int\limits\limits_{-1}^{-\frac{ 1 }{ 2 }}\sin^2(u) ( t^{-2} dt )\]

Answer 54

I don't know where to go from here. haha ;/

Answer 55

\[\large \color{green}{-du=t^{-2}dt}\]
We need to plug that piece in as well.
\[\large \int\limits\limits\limits_{-1}^{-\frac{ 1 }{ 2 }}\sin^2(u) (\color{green}{t^{-2} dt}) \qquad=\qquad \int\limits\limits\limits_{-1}^{-\frac{ 1 }{ 2 }}\sin^2(u) (\color{green}{-du})\]

Answer 56

then we antiderive right?

Answer 57

Yes, and unfortunately this one will be a little bit of a pain in the butt.
We'll need to apply the `Sine Half-Angle Formula`.

Answer 58

how does that work?

Answer 59

\[\large \sin^2x=\frac{1}{2}(1-\cos2x)\]

Answer 60

An important formula to remember.. especially for integration purposes :)

Answer 61

\[\large -\int\limits\limits_{t=-1}^{-1/2}\sin^2(u)\;du \qquad=\qquad -\frac{1}{2}\int\limits\limits_{t=-1}^{-1/2}1-\cos(2u)\;du\]

Answer 62

woahhhh... haha

Answer 63

Gotta remember these trig formulasssssss \:U/

Answer 64

of course. Just it look so complicated to evaluate now. haha

Answer 65

so now we plug the subs in and evaluate? or there's more to before applying the subs? @zepdrix

Answer 66

So we plug in our subs and evaluate :u ya.
Do you know how to find the anti-derivative of cos(2u)?

Answer 67

wouldn't it just be -sin(2u)

Answer 68

hmmmmmmmmmmmmmmm, no +_+

Answer 69

cause cos(x) = -sin(x) so going backwards -sin =cos?

Answer 70

@zepdrix

Answer 71

derivative of sine is cosine right?
So going backwards, integral of cosine is sine.
Positive :O not negative.

Also we have to deal with the fact that there is a 2 inside.

Answer 72

What is the derivative of sin(2u)?

Answer 73

2(cos(2u))

Answer 74

Hmm see that extra 2 on the outside?
It didn't give us cos(2u).
So our anti-derivative isn't quite right.

Answer 75

Mmmm anyway, here's something to keep in mind.
If we took the derivative of cos(2u), it produces an extra factor of 2.
When we integrate, we'll instead end up dividing by 2.\[\large \int\limits\limits \cos(2u)du \qquad=\qquad \frac{1}{2}\sin(2u)\]

Answer 76

okay, I understand now..

Answer 77

wait so..
\[\frac{ -1 }{ 2 } \int\limits_{-1}^{-\frac{ 1 }{ 2 }} 1-\cos(2u) du = 1-\frac{ 1 }{ 2 }\sin(2u)\]

Answer 78

@zepdrix

Answer 79

\[\large -\frac{1}{2}\int\limits\limits\limits_{t=-1}^{-1/2}1-\cos(2u)\;du \qquad=\qquad -\frac{1}{2}\left(?-\frac{1}{2}\sin(2u)\right)\]
Integral of 1? :o

Answer 80

1 or -1?

Answer 81

Im confused @zepdrix

Answer 82

\[=> -\frac{ 1 }{ 2 } (1-\frac{ 1 }{ 2 } \sin(2))\]
@zepdrix

Answer 83

\[\large \int\limits 1\;du \qquad=\qquad u\]

Answer 84

wait, how'd you get that? @zepdrix

Answer 85

Ah sorry the notifications aren't working :( hard to see your responses in a timely manner lol.

Site is acting up again -_- grumble grumble...

Answer 86

How did I integrate 1? is that the question?

Answer 87

It's okay haha. You've been a real help !(: and I really approciate it, so no worries. And I can agree on the site, it's lagging. And from the top how'd you get the last function you presented.

Answer 88

\[-\frac{1}{2}\int\limits\limits_{-1}^{-1/2}1-\cos(2u)\;du \qquad=\qquad -\frac{1}{2}\left[\int\limits\limits_{-1}^{-1/2}1\;du-\int\limits\limits_{-1}^{-1/2}\cos(2u)\;du\right]\]

Answer 89

okay, I got up to there too, but how you get
\[\int\limits_{}^{} 1 du = u\]

Answer 90

\[\large 1=u^0\]

\[\large \int\limits 1\;du \qquad=\qquad \int\limits u^0\;du\]Apply the power rule.
That will give us the first piece of our integral from above.

Answer 91

oh okay... So what's next? hah. I quite lost after this.

Answer 92

Integrating gives us,
\[\large -\frac{1}{2}\left[\int\limits\limits\limits_{-1}^{-1/2}1\;du-
\int\limits\limits\limits_{-1}^{-1/2}\cos(2u)\;du\right] \quad=\quad -\frac{1}{2}\left[u-\frac{1}{2}\sin(2u)\right]_{t=-1}^{1/2}\]

And then from here, we can either undo our substitution, or come up with new limits for our u.

Answer 93

\[\large u=1+\frac{1}{t}\]

\[\large t=-1 \qquad\to\qquad u=1+\frac{1}{-1} \qquad\to\qquad u=0\]
\[\large t=-1/2 \qquad\to\qquad u=1+\frac{1}{-1/2} \qquad\to\qquad u=-1\]
These will be the new limits on our u.\[\large -\frac{1}{2}\left[u-\frac{1}{2}\sin(2u)\right]_{t=-1}^{1/2} \qquad=\qquad -\frac{1}{2}\left[u-\frac{1}{2}\sin(2u)\right]_{u=0}^{-1}\]

Answer 94

\[-\frac{ 1 }{ 2 }\left[ 1-\frac{ 1 }{ 2 } \sin(2*1) \right] - \left( -\frac{ 1 }{ 2 } \right)\left[ 0-\frac{ 1 }{ 2 }\sin(2*0) \right] \]

Answer 95

That upper limit is -1, not 1 :o

Answer 96

ohhhh okay. gotcha.

Answer 97

I got
(-0.2273243567)-(0) so -0.2273243567 is the answer?