Recent content by aleksbooker1

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    MHB Reduction formula instead of using identities for trigonometric integration?

    This is one of the example problems in my book to show how to deal with integrating trigonometric functions to higher powers, by breaking them down into identities. $$=\int cos^5x dx$$ $$=\int (cos^2x)^2cos^x dx $$ $$=\int (1-sin^2x)^2*d(sin x)$$ $$=\int (1-u^2)^2 du$$ $$=\int 1-2u^2 + u^4...
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    MHB Stuck trying to integrate the surface area of a curve

    Thanks @MarkFL and @HallsofIvy, I'll add that to my mental cheatsheet for antiderivatives. :)
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    MHB Stuck trying to integrate the surface area of a curve

    Oh, okay. I know how to do that. The trick is anti-deriving with a variable in the exponent. For example, if I expand it and then split it into three integrands and try tackle each one by itself, I trip up on the first one: $$ e^{2y} $$. The best I can come up with is: $$ \frac{1}{2 \ln{2}...
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    MHB Stuck trying to integrate the surface area of a curve

    Hi @MarkFL, Thanks for the response. When you say expand the integrand, do you mean like this? S = $$\frac{1}{2}\pi \int e^{2y} + 2 + e^{-2y}$$ If so, how do I use the Fundamental Theorem of Calculus. I know what FTOC stands for but I never learned it as a thing by itself. I was taught how to...
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    MHB Stuck trying to integrate the surface area of a curve

    Here's the problem I was given: Find the area of the surface generated by revolving the curve $$x=\frac{e^y + e^{-y} }{2}$$ from 0 $$\leq$$ y $$\leq$$ ln(2) about the y-axis. I tried the normal route first... g(y) = x = $$\frac{1}{2} (e^y + e^{-y})$$ g'(y) = dx/dy = $$\frac{1}{2} (e^y -...
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