Query 4

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course mth 174

4/26 12:45 am

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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Question: Section 7.2 Problem 3

7.2.3 (previously 7.2.12. (3d edition 7.2.11, 2d edition 7.3.12)) Give an antiderivative of sin^2 x

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Your solution:

I used the trig identities to solve this integral. Since sin^2x is equal to (1-cos2x)/2 we take the integral of (1-cos2x)/2

So I let u = 2x

Du = 2dx

½ du = dx

So I rewrite as ½ times the integral (1-cosU)/2

½(x-sinu)/2

½(x-sin2x)/x+c

confidence rating #$&*: 3

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Given Solution: Good student solution:

The answer is -1/2 (sinx * cosx) + x/2 + C

I arrived at this using integration by parts:

u= sinx u' = cosx

v'= sinx v = -cosx

int(sin^2x)= sinx(-cosx) - int(cosx(-cosx))

int(sin^2x)= -sinx(cosx) +int(cos^2(x))

cos^2(x) = 1-sin^2(x) therefore

int(sin^2x)= -sinx(cosx) + int(1-sin^2(x))

int(sin^2x)= -sinx(cosx) + int(1) - int(sin^2(x))

2int(sin^2x)= -sinx(cosx) + int(1dx)

2int(sin^2x)= -sinx(cosx) + x

int(sin^2x)= -1/2 sinx(-cosx) + x/2

INSTRUCTOR COMMENT: This is the appropriate method to use in this section.

You could alternatively use trigonometric identities such as

sin^2(x) = (1 - cos(2x) ) / 2 and sin(2x) = 2 sin x cos x.

Solution by trigonometric identities:

sin^2(x) = (1 - cos(2x) ) / 2 so the antiderivative is

1/2 ( x - sin(2x) / 2 ) + c =

1/2 ( x - sin x cos x) + c.

note that sin(2x) = 2 sin x cos x.

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Self-critique (if necessary):

### I didn’t rewrite the sin2x - would my answer be acceptable or would points have been taken off if this were a test???

@&
For the given problem your solution would have been fine.

Be sure you also understand, as I expect you do, how integration by parts would have been used to get the equivalent answer.
*@

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Self-critique Rating:3

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Question: Section 7.2 Problem 4

problem 7.2.4 (previously 7.2.16 was 7.3.18) antiderivative of (t+2) `sqrt(2+3t)

**** what is the requested antiderivative?

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Your solution:

I begin with the integral of (t+2)(2+3t)^1/2

So I let u = t+2 v’=(2+3t)1/2

U’=1 V= 2/9(3t+2)^3/2

Now that we have the values of u, u’, v, v’ we use

U*v - integral(u’v)

(t+2)((2/9(3t+2)^3/2) - integral (1((2/9(3t+2)^3/2)

I have to take steps to find the integral of integral of {2/9(3t+2)^3/2}

The 2/9 comes to the outside of the integral so we have

2/9 *integral(3t+2)^3/2

U = 3t+2

Du=3dt

1/3du=dt

1/3(2/5u^5/2)

We have to multiply, 2/9*1/3* 2/5 which is 4/135 so we have

4/135(3t+2)^5/2 as the second part of the equation so rewriting we have

(t+2)((2/9(3t+2)^3/2) - 4/135(3t+2)^5/2

So I factor out (3t-2)^3/2

(3t-2)^3/2 {2/9(t+2)-4/135(3t+2)}+c

confidence rating #$&*: 2

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Given Solution: If you use

u=t+2

u'=1

v'=(2+3t)^(1/2)

v=2/9 (3t+2)^(3/2)

then you get

2/9 (t+2) (3t+2)^(3/2) - integral( 2/9 (3t+2)^(3/2) dt ) or

2/9 (t+2) (3t+2)^(3/2) - 2 / (3 * 5/2 * 9) (3t+2)^(5/2) or

2/9 (t+2) (3t+2)^(3/2) - 4/135 (3t+2)^(5/2). Factoring out (3t + 2)^(3/2) you get

(3t+2)^(3/2) [ 2/9 (t+2) - 4/135 (3t+2) ] or

(3t+2)^(3/2) [ 30/135 (t+2) - 4/135 (3t+2) ] or

(3t+2)^(3/2) [ 30 (t+2) - 4(3t+2) ] / 135 which simplifies to

2( 9t + 26) ( 3t+2)^(3/2) / 135.

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Self-critique (if necessary):

Ok!!! Yes, I got this one!! I feel like I am catching on more and more!

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Self-critique Rating:3

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Question: Section 7.2 Problem 8

**** problem 7.2.8 (previously 7.2.27 was 7.3.12) antiderivative of x^5 cos(x^3)

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Your solution:

Let u = x^3

Du=3x^2

V’= x^2cosx^3

To find v we have to find antiderivative of v’

So let u = x^3

Du=3x^2dx

1/3du=x^2

So 1/3 time integral of cosu^3 du

So finally v= 1/3sinx^3

Now we use u*v-integral(u’v)

X^3(1/3sinx^3)-integral 3x^2(1/3sinx^3)

Focusing on solving the integral first we see the 3 and 1/3 cancel giving us the integral of x^2sinx^3

Let u = x^3

Du=3x^2

1/3du=x^2dx

So 1/3 integral (sinu du)

-1/3 cos x^3

So now we have 1/3x^3sinx^3 - 1/3 cos x^3

We can factor out the 1/3 getting 1/3(x^3sinx^3 - cosx^3)+c

confidence rating #$&*: 3

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Given Solution:

It usually takes some trial and error to get this one:

• We could try u = x^5, v ' = cos(x^3), but we can't integrate cos(x^3) to get an expression for v.

• We could try u = cos(x^3) and v' = x^5. We would get u ' = -3x^2 cos(x^3) and v = x^6 / 6. We would end up having to integrate v u ' = -x^8 / 18 cos(x^3), and that's worse than what we started with.

• We could try u = x^4 and v ' = x cos(x^3), or u = x^3 and v ' = x^2 cos(x^3), or u = x^2 and v ' = x^3 cos(x^3), etc..

The combination that works is the one for which we can find an antiderivative of v '. That turns out to be the following:

Let u = x^3, v' = x^2 cos(x^3).

Then u' = 3 x^2 and v = 1/3 sin(x^3) so you have

1/3 * x^3 sin(x^3) - 1/3 * int(3 x^2 sin(x^3) dx).

Now let u = x^3 so du/dx = 3x^2. You get

1/3 * x^3 sin(x^3) - 1/3 * int( sin u du ) = 1/3 (x^3 sin(x^3) + cos u ) = 1/3 ( x^3 sin(x^3) + cos(x^3) ).

It's pretty neat the way this one works out, but you have to try a lot of u and v combinations before you come across the right one.

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Self-critique (if necessary):

### Do we have to put + c (I keep getting confused on whether or not to add the integration constant, I thought I was supposed to add it on general antiderivatives and integrals) I also tried a lot of trial and error and couldn’t get the u and v’ values, I ended up looking at what you used and understood with no problems, any tips on how to choose the u and v easier???

@&
When finding a general antiderivative you need the + c, and unless you're doing a definite integral it's always a good idea.

However if the question we just to find 'an antiderivative', then any antiderivative would do.

If the question asked you to find 'the antiderivative' then the + c should be included.

I just checked the problem and it's listed as an indefinite integral, so I should have included the + c in my given solution.
*@

@&
As far as finding possible u and v values, my best advice is to do a lot of integrals. Experience and practice are irreplaceable.

As illustrated in the given solution, you could try different powers of x for u. It wouldn't take much trial and error to arrive at u = x^3 as a possibility.
*@

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Self-critique Rating:3

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Question: **** What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?

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Your solution:

I tried a lot of trial and error without success, and once I realized what I needed to use for u and v’ I didn’t have any trouble. I integrated by parts. I used u substitution to solve for v’. I used u substitution again to solve the integral. Then I factored the 1/3 out of the final answer.

confidence rating #$&*: 3

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Given Solution:

TYPICAL STUDENT COMMENT:

I tried several things:

v'=cos(x^3)

v=int of v'

u=x^5

u'=5x^4

I tried to figure out the int of cos(x^3), but I keep getting confused:

It becomes the int of 1/3cosudu/u^(1/3)

I feel like I`m going in circles with some of these.

INSTRUCTOR RESPONSE:

As noted in the given solution, it often takes some trial and error. With practice you learn what to look for.

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00:53:03

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Self-critique (if necessary):

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Self-critique Rating:

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Question: Section 7.2 Problem 13

problem7.2.13 (previously 7.2.50 was 7.3.48) f(0)=6, f(1) = 5, f'(1) = 2; find int( x f'', x, 0, 1).

**** What is the value of the requested integral?

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Your solution:

I begin by letting u = x and u’=1

I let v’=f’(X) and v = f(x)

So now I use u*v-integral(u’v)

So I have xf’(x)-integral 1*f(x)

Xf(X)-integral f(x) the integral of f(x) is f(x) so we have

xf(x)-f(x)

we are told that f’(1)=2, F(1)=5 and f(0)=6

At this point I am confused on how to proceed, I

##??? I had the same steps as the given solution up until this point :

xf'(x)-integral of f'(x) ( I got this step)

Integral of f'(x) is f(x). So antiderivative is ( I got this step)

x f ' (x)-f(x) Why did the antiderivative become xf”(x) - f(x) instead of xf(x) - f(x)???

Besides this I am still having a hard time figuring the value of the interval. I think I should use the first fundamental theorem of calculus, and I know I evaluate at each value of the integral but I am confused. Could you walk me through this process and also if available provide me a formula like you did with the chain rule that would clarify for me???

confidence rating #$&*: 1

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Given Solution:

You don't need to know the specific function. You can find this one using integration by parts:

Let u=x and v' = f''(x). Then

u'=1 and v=f'(x).

uv-integral of u'v is thus

xf'(x)-integral of f'(x)

Integral of f'(x) is f(x). So antiderivative is

x f ' (x)-f(x), which we evaluate between 0 and 1. Using the given values we get

1 * f'(1)- (f(1) - f(0)) =

f ‘ (1) + f(0) - f(1) =

2 + 6 - 5 = 3.

STUDENT COMMENT: it seems awkward that the area is negative, so I believe that something is mixed up, but I have looked over it, and I`m not sure what exactly needs to be corrected

** the integral isn't really the area. If the function is negative then the integral over a positive interval will be the negative of the area. **

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**** Query Add comments on any surprises or insights you experienced as a result of this assignment.

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00:58:57

This was a very tedious assignment, but it will surely be a useful tool in computing areas over fixed integrals in the future. I do need more practice at these integrals, because I feel as if I`m going in circles on some of them. Any suggestions for proper techniques or hints on how to choose u and v? I have tried to look at how each variable would integrate the easiest, but I seem to make it look even more complex than it did at the beginning.

** you want to look at it that way, but sometimes you just have to try every possible combination. For x^5 cos(x^3) you can use

u = x^5, v' = cos(x^3), but you can't integrate v'. At this point you might see that you need an x^2 with the cos(x^3) and then you've got it, if you just plow ahead and trust your reasoning.

If you don't see it the next thing to try is logically u = x^4, v' = x cos(x^3). Doesn't work, but the next thing would be u = x^3, v' = x^2 cos(x^3) and you've got it if you work it through.

Of course there are more complicated combinations like u = x cos(x^3) and v' = x^4, but as you'll see if you work out a few such combinations, they usually give you an expression more complicated than the one you started with. **

This assignment was very time consuming because many of the problems had to be worked several times to achieve a

suitable answer. I will definitely need to practice doing more

** Integration technique does take a good deal of practice. There really aren't any shortcuts.

It's very important, of course, to always check your solutions by differentiating your antiderivatives. This helps greatly, both as a check and as a way to begin recognizing common patterns. **

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Supplemental questions:

If you are having trouble with this section you are invited to submit your solutions, including all steps and explanations, to the following.

If you wish to submit, just copy these questions into a text editor and insert your responses in the line preceding the #$&* mark.

Derivatives of composites:

You should be able to quickly calculate the following derivatives:

1. cos(e^x)

****

-sine^x*e^x

#$&*

2. sin(x^4 - 7 x^2)

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X^2(cos x^2-7)*2x(2x^2-7)

@&
The derivative would just be

(x^4 - 7 x^2) ' * cos( x^2 - 7 x^2)

= (4 x^3 - 14 x) * cos( x^2 - 7 x^2).

Notice that with the notation

(f(g(x))) ' = g ' (x) * f ' (g(x))

the argument of f ' (g(x)) is the same as that of f(g(x).

For this problem g(x) = x^2 - 7 x^2.

So both the original sine function and the cosine of the derivative will be applied to x^2 - y x^2.

Note also that the x^2 cant be factored out of

cos(x^4 - 7 x^2).

The cosine is not a linear function. In general, cos( a * b ) is not equal to a * cos(b).
*@

#$&*

****

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3. e^(x^3)

****

E^x^3*3x^2

#$&*

4. (cos(x))^4

****

-4x^3sinx^4

@&
There is no x^4 in this expression, so you won't get 4 x^3 in the derivative.

The functions are f(z) = z^4 and g(x) = cos(x).

f ' (z) = 4 z^3.
g ' (x) = -sin(x)

and the derivative is

-sin(x) * 4 (cos(x) ) ^ 3
= -4 sin(x) cos^3(x).

*@

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5. (e^(5x))^2

E^10x = 10e^10x

****

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6. sqrt(sin(x))

****

½(sinx)^-1/2 * cosx

@&
f(z) = sqrt(z) so f ' (z) = 1/2 z^(-1/2).

g(x) = sin(x) so g ' (x) = cos(x). There isn't an additional 1/2.

The derivative is

1/2 cos(x) (sin(x))^(-1/2).
*@

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Antiderivatives by simple substitution

You should be able to quickly and easily find antiderivatives of the following, all of which can be done using simple substitution. You should then be able to take the derivatives of your results to verify your antiderivatives:

7. e^(4 x )

****

U=4x

Du=4dx

1/4du=dx

Int. 1/4e^u

1/4e^4x+c

@&
Good.

You do want to be careful with signs of grouping, though.

1/4e^4x+c actually means

1/4 * (e^4) * x + c, not the correct (and intended)

1/4 e^(4x) + c.

You have other expressions with similar grouping omissions, and want to pay a little more attention to this.

However that's not the main point here so as long as I can tell what you mean, I'll limit my comments on this topic to this one problem.
*@

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8. sin ( x / 2 )

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U=x/2

Du=1/2dx

2du=dx

2* Int. sinudu

-2cosu

-2cosx/2+c

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9. ( 3 x - 4 ) ^ (3/2)

****

U=3x-4

Du=3dx

1/3du=dx

1/3 * integral u^3/2

1/3*2/5u^5/2

2/15(3x-4)^5/2+c

??? I think this one is incorrect when I check by taking the derivative, I cannot see where my mistake is???

#$&*

@&
I think this is correct, but if you can't get the derivative to match the original expression it would be a good idea to show me the steps you are following in taking the derivative.
*@

10. sqrt( 3 x - 4 )

****

U=3x-4

Du=3dx

1/3ddu=dx

1/3*integral u^3/2du

1/3*2/5u^5/2

2/15(3x-4)^5/2+c

@&
This doesn't seem to match the problem.
*@

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11. x e^(x^2)

****

U=x^2

Du=2xdx

1/2du=xdx

½*integral(e^udu)

1/2e^x^2+c

#$&*

12. y^2 cos(y^3)

****

U=y^3

Du=3y^2

1/3du=y^2

1/3*integral cosudu

1/3sinu

1/3 sin y^3+c

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Integration by parts using given breakdowns

You should be able to apply integration by parts to get the following, using the given information about u and v '

13. integral ( x e^x dx ) using u = x and v ' = e^x

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U=x u’=1 v’=e^x v=e^x

Xe^x-integral(1*e^x)

Xe^x-e^x+c

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14. integral of ( t cos(t) dt) using u = t and v ' = cos(t)

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U=t u’=1 v’=sint v=sint

Tsint-integral 1*sint

Tsint+cost+c

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15. integral of (y e^(4y) dy) using u = y and v ' = e^(4 y)

****

U=y u’=1 v’=e^4y v=1/5e^5y

Y(1/5e^5y)-integral(1*(1/5e^5y)

1/5ye^5y -1/5*1/5e^5y

1/5ye^5y-1/25e^5y

1/5e^5y(Y-1/5)+c

@&
An antiderivative of e^(4 y) is 1/4 e^(4 y).

If you take the derivative of e^(5 y) you get 5 e^(5 y); no 4y is present to match the original integrand.
*@

#$&*

16. integral of ( x ln(x) dx) using u = ln(x) and v ' = x

U=lnx u’=1/x v’=x v=(1/2x^2)

U*v-integral(u’*v)

Lnx*1/2x^2-intergral(1/x ((1/2)x^2)

1/2x^2lnx-1/2 *integral x^-1*x^2

1/2x^2lnx-1/2*integral x^1

1/2x^lnx-1/2((1/2)x^2)

1/2xlnx-1/4x^2

1/2x^2(lnx-1/2)+c

###I feel like I am off here somewhere but I’m not sure???

****

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17. integral of (t^2 sin(t^2) dt) using u = t and v ' = t sin(t^2)

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U=t u’=1 v’=tsintt^2 v=-1/2cost^2

T(-1/2cost^2)-integral(1*-1/2cost^2)

T(-1/2cost^2)-1/2*integral cost t^2

T(-1/2costt^2)-1/2*(1/2sint^2)

T(-1/2cost^2)-1/4sint^2

#$&*

18. integral of (x^5 cos(x^5) dx) using u = x and v ' = x^4 cos(x^5)

****

U=x u’=1 v’=x^4cosx^5 v=1/5sinx5 (found by using u substitution)

Now we have

X(1/5sinx^5)-integral(1*(1/5sinx^5)

Now I need to work on the right hand side

1/5*integral sinx(sinx^4)

1/5*integral sinx(sinx^2)^2

1/5*integral sinx(1-cosx)^2

Now I let u = 1-cosx so that du=sinxdx

So I have 1/5*integral u^2 du

1/5(1-cosx)^2

So now we add back on the left hand side

X(1/5sinx^5)- 1/5(1-cosx)^2 +c

X(1/5sinx^5)-1/5(sinx)^2+c

### I think I must have made a mistake somewhere because if I continue on I am going to get zero???

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Figuring out possible breakdowns into u and v ', given u or v '

You should be able to apply integration by parts to the following, where you are given either u or v ':

19. integral of ( x^2 e^(x^2) dx ) using u = x

****

U=x u’=1 v’=xe^x^2 v= 1/2e^x^2

X(1/2e^x^2)-1/2 *integrale^x^2

X(1/2E^x^@)-1/2(2e^x^2)

X(1/2E^x^@)-e^x^2

E^x^2(1/2X-1)+c

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20. integral of ( t^4 cos( t^4) dt) using u = x

****

###I wasn’t sure how to proceed with this one using u=x I am confused as to where the x came from????

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21. integral of (y^4 e^(y^4) dy) using v ' = y^3 e^(y^4)

****

U=y u’=1 v’=y^3e^y^4 v=1/4e^y^4 (use u substitution to find v)

Y(y^3E^y^4)*integraly^3e^y^4 (use u substitution to find integral)

Y(y^3e^y^4)*1/4eY^4

E^y^4(Y^4-1/4)+c

#$&*

22. integral of ( 1/3 t^2 sin( 1/4 t^2)dt) using v ' = t sin( 1/4 t^2)

****

U=1/3t

U’=1/3

V’=tsin(1/4t^2)

V=-1/2cost^2

1/3t(-1/2cost^2)*integral 1/3(-1/2cost^2)

1/3t(-1/2cost^2)*-1/6*integral cost^2

-1/6tcostt^2*-1/6(-2tsint^2)

-1/6tcost^2*1/3tsingt^2

1/3t(-1/3cost^2*sint^2) +c

#$&*

Self-critique (if necessary):

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Self-critique rating:

@&
Good work, but be sure to see my notes.

You proably need a little more practice on taking derivatives. Your integration technique looks good; your derivatives aren't bad but are sometimes a little shaky.

Check my notes and let me know if you have questions.
*@

Query 4

@& For the given problem your solution would have been fine. Be sure you also understand, as I expect you do, how integration by parts would have been used to get the equivalent answer. *@

@& As far as finding possible u and v values, my best advice is to do a lot of integrals. Experience and practice are irreplaceable. As illustrated in the given solution, you could try different powers of x for u. It wouldn't take much trial and error to arrive at u = x^3 as a possibility. *@

@& There is no x^4 in this expression, so you won't get 4 x^3 in the derivative. The functions are f(z) = z^4 and g(x) = cos(x). f ' (z) = 4 z^3. g ' (x) = -sin(x) and the derivative is -sin(x) * 4 (cos(x) ) ^ 3 = -4 sin(x) cos^3(x). *@

@& f(z) = sqrt(z) so f ' (z) = 1/2 z^(-1/2). g(x) = sin(x) so g ' (x) = cos(x). There isn't an additional 1/2. The derivative is 1/2 cos(x) (sin(x))^(-1/2). *@

@& I think this is correct, but if you can't get the derivative to match the original expression it would be a good idea to show me the steps you are following in taking the derivative. *@

@& This doesn't seem to match the problem. *@

@& An antiderivative of e^(4 y) is 1/4 e^(4 y). If you take the derivative of e^(5 y) you get 5 e^(5 y); no 4y is present to match the original integrand. *@

@& Good work, but be sure to see my notes. You proably need a little more practice on taking derivatives. Your integration technique looks good; your derivatives aren't bad but are sometimes a little shaky. Check my notes and let me know if you have questions. *@

@&
For the given problem your solution would have been fine.

Be sure you also understand, as I expect you do, how integration by parts would have been used to get the equivalent answer.
*@

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As far as finding possible u and v values, my best advice is to do a lot of integrals. Experience and practice are irreplaceable.

As illustrated in the given solution, you could try different powers of x for u. It wouldn't take much trial and error to arrive at u = x^3 as a possibility.
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There is no x^4 in this expression, so you won't get 4 x^3 in the derivative.

The functions are f(z) = z^4 and g(x) = cos(x).

f ' (z) = 4 z^3.
g ' (x) = -sin(x)

and the derivative is

-sin(x) * 4 (cos(x) ) ^ 3
= -4 sin(x) cos^3(x).

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f(z) = sqrt(z) so f ' (z) = 1/2 z^(-1/2).

g(x) = sin(x) so g ' (x) = cos(x). There isn't an additional 1/2.

The derivative is

1/2 cos(x) (sin(x))^(-1/2).
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I think this is correct, but if you can't get the derivative to match the original expression it would be a good idea to show me the steps you are following in taking the derivative.
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This doesn't seem to match the problem.
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An antiderivative of e^(4 y) is 1/4 e^(4 y).

If you take the derivative of e^(5 y) you get 5 e^(5 y); no 4y is present to match the original integrand.
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Good work, but be sure to see my notes.

You proably need a little more practice on taking derivatives. Your integration technique looks good; your derivatives aren't bad but are sometimes a little shaky.

Check my notes and let me know if you have questions.
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