Calculus Midterm...Please Help!!!

[lordtyranus2]

The following 3 integrals are to be on midterms tomorrow, and I cannot do them

int(cos(x)^4)



int(sec(x)^3*tan(x)^3



int((2x-1)/(x^2+x))



Please help, I will be gratified. I need solutions, meaning showing work. I already have answers.

 

[ngvepforever2]

I can help you, but you have to try it first. Otherwise you wouldn't learn anything.

Regards

 

[BigJ]

you ever hear of U-substitution?

 

[parsley007]

midterms already?? i'm in the third week of class... oh, and btw, calc101 has helped me before with derivatives, but you have to pay for the integral solutions (they're step by step with explanation).

 

[esun]

Try This

If you want all the steps you'll need to buy a password, though. There may be some free service that is similar.

 

[Soccer55]

Please help, I will be gratified.

Is anyone else disturbed by the wording here? 😛

At any rate, you should post what work you have done so that instead of just feeding you answers, we might be able to show you what you have and/or haven't done right, then you can work with the hints that are given to you in order to finish the problems.

-Tom

 

[EmperorIQ]

Originally posted by: ngvepforever2
I can help you, but you have to try it first. Otherwise you wouldn't learn anything.

Regards

i concur

 

[phatj]

For the first two, use apply trig identities. The last one is really easy u-sub.

 

[Ionizer86]

Give you some hints: Remember that sin(x)^2+cos(x)^2 = 1, and try some subbing. There's a similar formula for Tan and Sec. Look it up.

 

[lordtyranus2]

Me and 5 friends have tried integration by parts, u substitutions, etc etc. For hours.

Edit: I have gotten the sec tan one. Convert it all to sines and cosines, change a sin2x to 1-cos2x, and use a u substitution of cos x.

 

[BigJ]

Originally posted by: lordtyranus2
Me and 5 friends have tried integration by parts, u substitutions, etc etc. For hours.

Edit: I have gotten the sec tan one. Convert it all to sines and cosines, change a sin2x to 1-cos2x, and use a u substitution of cos x.

No offense, but if it has taken you that long on these 3 problems, with text books, calculators, and all, you do not deserve to be in your current Calculus class. You also deserve to fail the mid-term because obviously you either lack the skill or have been fvcking around in class.

 

[PowerMacG5]

That last one is disgustingly easy.

EDIT: So is the first. The only one that could even give you a slight problem is the second, and that isn't even that hard. Like they said, if it took you this long with all the solutions at hand (text, calculators, internet, etc...) then you really should take an easier class.

 

[Goosemaster]

Originally posted by: BigJ

Originally posted by: lordtyranus2
Me and 5 friends have tried integration by parts, u substitutions, etc etc. For hours.

Edit: I have gotten the sec tan one. Convert it all to sines and cosines, change a sin2x to 1-cos2x, and use a u substitution of cos x.

No offense, but if it has taken you that long on these 3 problems, with text books, calculators, and all, you do not deserve to be in your current Calculus class. You also deserve to fail the mid-term because obviously you either lack the skill or have been fvcking around in class.

I wouldn't got that far...


wait...

*reads problems*


WOW! READ YOUR BOOK.
#1 is DEFINITELY IN EVERY CALCULUS BOOK EVER MADE


1. We ALL assume you are a lazy little bitch that wants everything done for him because he realized he has an exam tommorow and know he's otherwise royall Fvked.
2. Even so, if you post "3 hours " worth of work, I will gladly correct you.

 

[Goosemaster]

Originally posted by: KraziKid
That last one is disgustingly easy.

EDIT: So is the first. The only one that could even give you a slight problem is the second, and that isn't even that hard. Like they said, if it took you this long with all the solutions at hand (text, calculators, internet, etc...) then you really should take an easier class.

True dat.

Please, if you want LEGITIMATE HELP, regardless of your situation, post your work.

 

[lordtyranus2]

Ok, I'll start posting as best as I can.

The 3rd one looked easy at first, due to a du/u substitution. But as the numerator is 2x-1, not 2x+1, it does not work. I then continued to try adding one and subtracting 1 to the numerator. Then you get 2 integrals, 2x+1/x^2+x and -2/X^2+x. The first integral is ez, the second I cannot do anything with. I still do not have a solution to that one.

The second I ahve gotten. I did not realize that sin^3x should be split into sin^2x * sinx, then sin^2x converted via identity, then integration by substitution with u=cosx.

The cos^4x I have not gotten anywhere on. I have tried integration by parts with cos x, cos^2x, cos^3x. Just makes a bigger mess, and I cant integrate that bigger mess without a calculator.

WOW! READ YOUR BOOK.
#1 is DEFINITELY IN EVERY CALCULUS BOOK EVER MADE

Not in ours. I already checked. Well its in the table of integrals, but I need a solution, not only answer.

 

[lordtyranus2]

you ever hear of U-substitution?

Yes. To everyone who keeps saying this is a very easy problem.

READ IT AGAIN. It is NOT a simple u substitution. Look at the numerator, it is 2x-1, not 2x+1.

 

[Kermy]

Uh did you even try partial fractions on the third one?
A(x+1) + B(x) = 2x-1

 

[Alphathree33]

Originally posted by: Kermy
Uh did you even try partial fractions on the third one?
A(x+1) + B(x) = 2x-1

You dont need partial fractions. Just divide the denominator into each term of the numerator

You get
2/(x + 1) - 1/[(x + 1/2)^2 -1/4]

after completing the square on the second fraction

Now you can substitute for u = x + 1/2 in the second guy

for the first integral, substitute u = x + 1

 

[lordtyranus2]

We don't know partial fractions. I have an idea of how they work, but we have not covered them yet.