Recent content by Zack88
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Evaluating the integral, correct?
k cya thanks again- Zack88
- Post #116
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
lol so in other words was the square already completed?- Zack88
- Post #114
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
wouldnt the 9 - 9 cancel out? so ur still left with the original problem.- Zack88
- Post #112
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
ok came across a bump. [integral] dx/ x^2 - 6x + 13 so i have to complete the square, so x^2 - 6x + 13 = 0 x^2 - 6x = -13 wait before i continue I always thought anything that was negative and squared it become positive but now my calculator is telling me differently. the...- Zack88
- Post #110
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
oh ok and i just made up the example so if i have a number higher than 1 I need to divide that number out so ex. 5x^2 + 6x + 5 and then id divide the whole thing by 5.- Zack88
- Post #108
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
I don't know what that means, since i work with the square root do I not pay attention to it until finally getting the final answer. ex. x^3 + 6x^2 - 4 then x^3 + 6x^2 = 4 x^3 + 6x^2 + 9 = 4 + 9 and so on- Zack88
- Post #106
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
more like 1 / 9x^2 + 6x -8- Zack88
- Post #104
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
hey when completing the square when a square roots is involved, do you square both sides?- Zack88
- Post #102
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
sorry if i was irritating you it has been a year since I've done any math and I am jumping into calc 2, but i finally got it and I have done most of the rest, just three left which I am working on now, so thank you for all your help and putting up with me :)- Zack88
- Post #100
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
[integral] x/2 (1 + cos2x) [chain rule] [integral] x/2 (-2sin2x) + (1 + cos2x)(1/2) = x^2 /4 (cos2x) + (((sin(2x) / 2) + x)) (x/2) = (x^2 cos 2x) / 4 + x(sin2x / 2 + x/2)- Zack88
- Post #98
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
ok [integral] x cox^2 x dx turns into [integral] x/2(1 + cos2x) dx- Zack88
- Post #97
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
[integral] x cox^2 x dx u= x du = dx dv= cos^2 v= (sin(2x)) / 4 + x/2 x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c i know the answer is x (sin(2x)) / 4 + cos(2x) / 8 + x^2 / 4 + c but I got an extra x/2- Zack88
- Post #95
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
thank you must have wrote it down wrong on paper, but does that matter since i didnt use that identity?- Zack88
- Post #93
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
i know the answer is close to that except for the extra x/2- Zack88
- Post #91
- Forum: Calculus and Beyond Homework Help
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Evaluating the integral, correct?
[integral] x cox^2 x dx u= x du = dx dv= cos^2 v= (sin(2x)) / 4 + x/2 x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c- Zack88
- Post #90
- Forum: Calculus and Beyond Homework Help