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Loren's Ongoing Calculus Confusion
Loren Michael
Registered User regular
Registered User regular
Okay, this is based paritally on this one that I posted previously.
Basically, I get to ln y and then... shit, how do I integrate ln y? What's next? Is it 1/y? The answer I'm supposed to get is that the primitive function of ln y is ln y again, but I'm under the impression that that's wrong.
What's up with that?
Next, could someone please walk me through the integration of (9-y^2)^(1/2)? Originally, I got [(9-y^2)^(3/2)]/[-3y] but I am now under the impression that I am wrong, given that [(9-y^2)^(3/2)]/[-3y] doesn't differentiate back to (9-y^2)^(1/2). What's up with that? Where am I going wrong?
Finally, similar problem with (6-y)^(1/2). My integration is weak. :x
Basically, I get to ln y and then... shit, how do I integrate ln y? What's next? Is it 1/y? The answer I'm supposed to get is that the primitive function of ln y is ln y again, but I'm under the impression that that's wrong.
What's up with that?
Next, could someone please walk me through the integration of (9-y^2)^(1/2)? Originally, I got [(9-y^2)^(3/2)]/[-3y] but I am now under the impression that I am wrong, given that [(9-y^2)^(3/2)]/[-3y] doesn't differentiate back to (9-y^2)^(1/2). What's up with that? Where am I going wrong?
Finally, similar problem with (6-y)^(1/2). My integration is weak. :x
Loren Michael on
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The integral of ln(y) dy is a little tricky. Assuming you know integration by parts, we take one part to be ln(y) and the other part to be 1 dy. Then Integral(ln(y) dy) = y*ln(y) - Integral(1/y * y dy) = y*ln(y) - Integral(1 dy) = y*ln(y) - y.
As for (9-y^2)^(1/2): The best way to approach this is trigonometric substitution. We take y = 3*sin(t), then dy = 3*cos(t) dt, so we get:
Integral( (9 - 9*sin^2(t))^1/2) * 3*cos(t) dt)
= Integral( (9*cos^2(t))^1/2) * 3*cos(t) dt)
= Integral( 3*cos(t) * 3*cos(t) dt)
= 9 * Integral(cos^2(t) dt)
If you know the integral of cos^2 it's fairly simple now, otherwise you have to follow something like this to complete the problem.
what derivative gives you LN(x) as an answer.
well derivative of LN(x) is 1/x... derivative of xLn(x) with the chain rule gives you ln(x) + 1. so the question is "the derivative of what is -1" and the answer is -x. so from all that you get integral Ln(x) = x*Ln(x) - x
If you're familiar with regular substitution -- "u substitution", as it is frequently called in intro calc classes -- you can pick up trigonometric substitution quickly. Basically trigonometric substitution is the opposite of regular substitution. With regular substitution, we define u = f(x); with trigonometric substitution we define x = f(t), where f is typically a simple trigonometric substitution. Then you just do simple algebraic substitution, replacing every instance of x with f(t), and replacing dx with f'(t) dt.
As Iroh said, the reason it's the best approach for this particular problem is because of trig identities. In particular, you have 9 - x^2; if you take x = 3*cos(t) or x = 3*sin(t), you can use the old sin^2(t) + cos^2(t) = 1 to simplify the portion of your integrand which is inside the square root, which will in turn greatly simplify the whole problem.