Indefinite Integrals (Antiderivatives)

Hi, Kristenx2 again. Same take-home test.

I missed the day we did antiderivatives, but is seems easy enough... I hope.

My only issue is when it comes to the chain rule... Trig functions screw me up.

I would just appreciate someone checking my work for the first, and possibly showing me how to do the second. I know I should use the quotient rule, but really.

*Since I can't make that fancy antiderivative squiggle line, I will substitute with / *

(a) /(3sec*x*tan*x*+7sec^2*x*)*dx*

=3/2sec^2*x*+7tan*x*+c

(b) /(3*x*^3+5*x*^7)

--------------------------- *dx*

(square root of *x*)

= ? :(

Re: Indefinite Integrals (Antiderivatives)

Re: Indefinite Integrals (Antiderivatives)

Oh! Thank you for your help! As you can see, I'm horribly confused, so I'm very unsure as to where anything comes in.

I was looking on other sites as to how to do antiderivatives when there is a quotient, and they just said to flip it up.

So I have

/(3x^3+5x^7)(x^-1/2)dx.

And ended up with

6/7x^7/2+1/3x^15/2+c. I just used the power rule. I can use that even when there are constants, right?

Re: Indefinite Integrals (Antiderivatives)

I think you really need a catch up session with your teacher. Also, for part b) I did most of the work for you. It should be Did you manage to get anywhere with part a)?

Re: Indefinite Integrals (Antiderivatives)

Me, too. I noticed you got 2/3 where I got 1/3, that's probably just a mistake in my math.

For the first one, would it just be

3sec^2x+7tanx+c?

Just leave the constants and figure out what makes the trig functions?

Re: Indefinite Integrals (Antiderivatives)

No, did you try what I suggested in post #2?

Re: Indefinite Integrals (Antiderivatives)

Yes, I got sec(x)tan(x)+sec^2(x), which does look an awful lot like the original problem!

However, how do I get the coefficients to come into play?

Re: Indefinite Integrals (Antiderivatives)

The coefficients don't change, as we have:

Re: Indefinite Integrals (Antiderivatives)

Re: Indefinite Integrals (Antiderivatives)

Is a take-home test meant to be your own work?

Thread closed. See rule #6: http://www.mathhelpforum.com/math-he...hp?do=vsarules.

Thread closed.

Quote:

Originally Posted by

**Kristenx2** Hi, Kristenx2 again. Same take-home test.

I missed the day we did antiderivatives, but is seems easy enough... I hope.

My only issue is when it comes to the chain rule... Trig functions screw me up.

I would just appreciate someone checking my work for the first, and possibly showing me how to do the second. I know I should use the quotient rule, but really.

*Since I can't make that fancy antiderivative squiggle line, I will substitute with / *

(a) /(3sec*x*tan*x*+7sec^2*x*)*dx*

=3/2sec^2*x*+7tan*x*+c

(b) /(3*x*^3+5*x*^7)

--------------------------- *dx*

(square root of *x*)

= ? :(