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) /(3secxtanx+7sec^2x)dx
=3/2sec^2x+7tanx+c
(b) /(3x^3+5x^7)
--------------------------- dx
(square root of x)
= ?
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?
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?
This is just a matter of personal preference, but I don't like having to remember lots of integration formulas. For trigonometric functions I really only ever remember and . Every other time I'll try to convert the integrand into some combination of these...
Now make the substitution and the integral becomes
which is the same answer you would get using the other method as well, seeing as as you found before
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.