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Calculus and Beyond Homework Help
Integration by Parts: Does the Choice of u and dv Matter?
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[QUOTE="mrg, post: 4964170, member: 509673"] [B]Homework Statement [/B] $$ \int x^{3}cos(x^{2})dx$$ [B]The attempt at a solution[/B] OK, so I am aware that there is a way in which to do this problem where you do a substitution (let $$u=x^{2}$$ to do a substitution before you integrate by parts), and I was able to get the answer right using this method. The thing is, I tried it a different way first, and after triple checking it, I feel like I should have gotten it right. It doesn't matter what I choose to let [I]u[/I] and [I]dv[/I] equal, right? I should get the right answer no matter what I choose (assuming I did things correctly, of course). Here's my work: $$u=cos(x^{2})$$ $$du=-2xsin(x^{2})$$ $$v=\frac{x^{4}}{4}$$ $$dv=x^{3}dx$$ $$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}\int x*sin(x^{2})dx$$ $$u_2=sin(x^{2})$$ $$du_2=2xcos(x)$$ $$v_2=\frac{x^{2}}{2}$$ $$dv_2=x$$ $$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}(\frac{x^{2}sin(x^{2})}{2}-\int x^{3}cos(x^{2})dx)$$ Since this last integral is the same as the one we started with, we can now say that $$I=A+B-\frac{1}{2}I$$ Hence $$\frac{3}{2}I=A+B$$ Multiplying both sides by two over three... $$\frac{x^{4}cos(x^{2})}{6}+\frac{x^{2}sin(x^{2})}{6}$$Thanks for your time and any help you can offer me. [/QUOTE]
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Integration by Parts: Does the Choice of u and dv Matter?
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