1. Why does the derivitive of y= (cos^2(x) + sin^2(x))^6 equal 0? I don't see how it turns out that way. 2. Why does the integral of (1/(x+3)^3 with the lower boundary 2 and upper 5 equal 39/3200? 3. Finally, I don't understand how to integrate: [sq rt(ln x)/x] because I can't figure out how to make it work with the F'(x)/F(x) rule. Thanks, I just need to understand these 3 before the exam tomorrow!
First two are really easy: 1.[tex] (\cos^{2}x+\sin^{2}x)^{6}=1^{6}=1[/tex] [tex](1)'=0[/tex] 2.[tex] \int_{2}^{5} (\frac{1}{x+3})^{3} dx=\int_{2}^{5} (x+3)^{-3} dx[/tex] Make substitution [tex] x+3\rightarrow u[/tex] The new limits of integration are [tex] 2\rightarrow 5[/tex] & [tex]5\rightarrow 8[/tex] The new integral is [tex]\int_{5}^{8} u^{-3} du [/tex] which i hope you can show it yields [itex] \frac{39}{3200} [/itex] 3.I didn't understand.Is it the integral: [tex] \int \sqrt{\frac{\ln x}{x}} dx [/tex] ?????? If so,lemme think about it a little more. Daniel. PS.For number 1) i used the fundamental identity for the circular trigonometric functions.I hope it's not a stranger to u. :tongue2:
Yeah, for number 1, I understand, I just didn't think that sin^2 + cos^2 was 1. For the second one, how do you know that the limits change and what they change to. And for the last, the denominator is not under the radical sign.
If it's [tex] \int \frac{\sqrt{\ln x}}{x} dx [/tex] ,then it's an easy one. U know that (i hope :tongue2: ) [tex] d(\ln x)=\frac{dx}{x} [/tex] It follows immediately that your integral can be written [tex] \int \sqrt{\ln x} d(\ln x) [/tex] ,which can be easily computed making the substitution [tex] \ln x\rightarrow u [/tex] Daniel.
1. What do you know about the expression [itex]\cos^2x+\sin^2x[/itex]? (Some kind of identity maybe?) 2. Just integrate. Post what you have so far. Notice that [itex]\frac{1}{(x+3)^3}=(x+3)^{-3}[/itex] 3. What's the derivative of [itex](\ln x)^n[/itex] for any n?. EDIT: Darn, beaten by 3 posts.
The limits of integration need to change as u make change of the integration variable.You integrate after "x" which goes from "2" to "5".U define "u=:x+3" and integrate after "u".It's natural that the limits of integration (initially after "x") need to change,and they do according to your definition. Think the definition as a function of "x": [tex] u(x)=:x+3 [/tex] Then: [tex] u(2)=:2+3=5;u(5)=:5+3=8 [/tex] You know that the definite integral for one variable functions has a geometric meaning:they are the area under the graph of the integrated function,area bordered by the graph,Ox axis and two vertical straight lines x=x_{1} & x=x_{2},where x_{1} & x_{2} are the two limits of integration. Think of a change of variable for integration as a reparametrization for the function integrated: [tex] y(x)\rightarrow y(u(x)) [/tex] So,the new function will depend explictely of the new parameter "u".And so the limits of integration change naturally.Instead of x_{1} it will be u(x_{1}) and the same for x_{2}; Changes of variables of integration make integrals easier,but there are,as i showed,geometric foundation behind it.Not only algebraic. Daniel.