Recent content by calisoca
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Find the general indefinite integral
Thank you very much. I'm at that point where it's obvious this makes sense. If I hadn't been up for the last 48 hours, then I wouldn't even need to ask the question! Ha. Nonetheless, yes, I see superficially that this will make sense later, but my brain will not let me comprehend it...- calisoca
- Post #7
- Forum: Calculus and Beyond Homework Help
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Find the general indefinite integral
I apologize. It's been a very long day, and I'm crazy tired. I meant to give you the derivative. The problem is that we haven't done anything like what you just described, so I'm having a hard time figuring it out. It looks like you used some sort of substitution. I've looked ahead, and...- calisoca
- Post #5
- Forum: Calculus and Beyond Homework Help
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Find the general indefinite integral
\sin{x} + C I apologize, but I'm still not following.- calisoca
- Post #3
- Forum: Calculus and Beyond Homework Help
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Find the general indefinite integral
Homework Statement I wasn't really sure where to post this, as I don't need help understanding the integration. I need help with the trigonometry! That being said, here is the problem. Find the general indefinite integral of \int\frac{\sin{x}}{1-sin^2{x}}dx Homework Equations...- calisoca
- Thread
- General Indefinite Indefinite integral Integral
- Replies: 6
- Forum: Calculus and Beyond Homework Help
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Beginning Trignometric Antiderivative
Great. Thanks to everyone for their help!- calisoca
- Post #5
- Forum: Calculus and Beyond Homework Help
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Beginning Antiderivative Question
Yes, of course! Thank you, HallsofIvy.- calisoca
- Post #8
- Forum: Calculus and Beyond Homework Help
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Beginning Trignometric Antiderivative
Ah, crap! I don't know where I got \frac{1}{\cos^2{\theta}}} \ = \ \csc^2{\theta} from. That's obviously not right. Thank you for pointing out my stupid mistake. So... 1.) f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}} 2.) f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ +...- calisoca
- Post #3
- Forum: Calculus and Beyond Homework Help
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Beginning Antiderivative Question
Thank you very much everyone for all of your help!- calisoca
- Post #6
- Forum: Calculus and Beyond Homework Help
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Beginning Trignometric Antiderivative
Homework Statement Okay, I think I'm finally getting the hang of these antiderivatives. However, I'm still stumbling some on trigonometric functions. Find the antiderivative of f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}} Homework Equations f(\theta) \ = \ \frac{1 +...- calisoca
- Thread
- Antiderivative Beginning
- Replies: 4
- Forum: Calculus and Beyond Homework Help
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Beginning Antiderivative Question
Well, I know that, for example, if asked to find the antiderivative of 5, the answer would be 5x + C . I was trying to use similar logic here by saying 5^\frac{1}{2} would become something like \frac{10\sqrt{x^3}}{3} . However, that didn't seem right. So, if I follow the logic that...- calisoca
- Post #4
- Forum: Calculus and Beyond Homework Help
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Beginning Antiderivative Question
Homework Statement Find the antiderivative of F(x) \ = \ \sqrt{\frac{5}{x}} Homework Equations F(x) \ = \ \sqrt{\frac{5}{x}} The Attempt at a Solution 1.) F(x) \ = \ \sqrt{\frac{5}{x}} 2.) F(x) \ = \ (\frac{5}{x})^\frac{1}{2} 3.) F(x) \ = \ \frac{5^\frac{1}{2}}{x^\frac{1}{2}}...- calisoca
- Thread
- Antiderivative Beginning
- Replies: 7
- Forum: Calculus and Beyond Homework Help
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Massively complex anti-derivative. Impossible?
Yea, I'm starting to see it now. However, if you hadn't mentioned it, I probably never would have seen it! Ha! Anyway, thanks for the pointer. I'm working on it, and I'm slowly getting there. Thanks.- calisoca
- Post #3
- Forum: Calculus and Beyond Homework Help
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Massively complex anti-derivative. Impossible?
Homework Statement Find the anti-derivative of the following equation. Homework Equations \frac{df}{dx} = \frac{[\frac{(30x^2 + 10x + 3)(\sqrt[3]{\frac{(4x^3 + 2x^2)}{5x^2}})}{(5)\sqrt[5]{(\frac{(10x^4 + 5x^3 + 3x^2)}{6x})^4}}] \ - \ [\frac{(\frac{(60x^4 - 20x -...- calisoca
- Thread
- Complex Impossible
- Replies: 2
- Forum: Calculus and Beyond Homework Help
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From point A to point B, basic algebra.
Crap! Just figured it out. Like Danago said, \frac{a^2}{b^2} = (\frac{a}{b})^2 So... 1.) \frac{81}{n^4} \ [\frac{n(n+1)}{2}]^2 \ - \ \frac{54}{n^2} \ [\frac{n(n+1)}{2}] 2.) \frac{81}{n^4} \ [\frac{n^2 + n}{2}]^2 \ - \ \frac{54}{n^2} \ [\frac{n^2 + n}{2}] 3.)...- calisoca
- Post #5
- Forum: Precalculus Mathematics Homework Help
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From point A to point B, basic algebra.
Okay, here is what I have so far. However, getting the first term to what I want is problematic for me. Step 6 is where I am stuck. May I please get some help getting past Step 6? I'd greatly appreciate it.1.) \frac{81}{n^4} \ [\frac{n(n+1)}{2}]^2 \ - \ \frac{54}{n^2} \ [\frac{n(n+1)}{2}]...- calisoca
- Post #4
- Forum: Precalculus Mathematics Homework Help