I really need help with some problems from my Math HW.

I need to find the most general antiderivative:


#3)

3w^5 - 2w^3 - 4
--------------------------
w^3


#7)

cos^8(pheta) * sin(pheta)




#14)

(1-y)
---------------
(1 + y^2)




#32)

3t-15
----------------------------------------
cube root of (t^2 -10t + 1)




#45)

arctanx
------------ On the integral: negative pie/3 to positive pie/2
1 + x^2

nm

This post was edited by JDota72 on Sep 6 2014 04:38pm

user expects fully shown work and won't understand basic explanations. told me he would pay me small amounts of fg for help and then ended up not paying, saying my work was incorrect.
don't recommend helping him.

This post was edited by JDota72 on Sep 6 2014 10:26pm

check your inbox.

To clear this up, I stated from start I would need work shown so i could understand the problem and learn from it. You stated I expect fully shown work, you showed no work.
I said I would give you a donation of 250fg for answer with work clearly shown. And I gave you a donation of 125fg for only the answers (of which the answers do not help me at all). Answers were incorrect and explanations were not good enough for my knowledge of Calc (hence why I asked for full work shown).

I was very nice and polite, Im sorry we had this misunderstanding.

And again, check your inbox. I have an idea that may better the situation for both of us

k ill give it another try

try
http://www.wolframalpha.com/

I think you get 3 complete free solutions per day

Different rules should be applied for each antiderivative. C denotes a constant.

#3 Cut your function into pieces :

( 3w^5 - 2w^3 -4) / w^3 = 3w^2 - 2 - 4/w^3

Take an antiderivative of each part :

w^3 - 2w + 2/w^2 + C


#7 Observe that sin is (almost) the derivative of cos and use the power rule : (u^n) ' = n.u'.u^(n-1)

( cos ^ 9) ' = 9.(-sin).cos^8


#14 Cut the fraction into 2 pieces, and observe that a numerator is (almost) the derivative of a denominator.
Use the rule for Ln : (Ln u) ' = u' / u and remember that (Arctan y)' = 1 / (1+y²)

In the end, a general anti-derivative is Arctan y - 0.5 * Ln (1+y²) + C


#32 Observe that the derivative of t² - 10t +1 is almost your numerator, and use the power rule with n = 2/3 :

(t² -10t +1) ^ (2/3) = (2/3).( 2t - 10).(t² -10t +1)^(-1/3)

Remember that X ^ (-1/3) = 1 / cube root of X

#45 Your function is of the form u'.u, with u = Arctan.

Remember that (u²) ' = 2.u'.u (the power rule again)

Feel free to pm back if something is not clear.

Good luck !