# Differentiation and Integration questions

by Sombra
Tags: differentiation, integration
 Sci Advisor HW Helper P: 11,927 First two are really easy: 1.$$(\cos^{2}x+\sin^{2}x)^{6}=1^{6}=1$$ $$(1)'=0$$ 2.$$\int_{2}^{5} (\frac{1}{x+3})^{3} dx=\int_{2}^{5} (x+3)^{-3} dx$$ Make substitution $$x+3\rightarrow u$$ The new limits of integration are $$2\rightarrow 5$$ & $$5\rightarrow 8$$ The new integral is $$\int_{5}^{8} u^{-3} du$$ which i hope you can show it yields $\frac{39}{3200}$ 3.I didn't understand.Is it the integral: $$\int \sqrt{\frac{\ln x}{x}} dx$$ ?????? 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.
 Sci Advisor HW Helper P: 11,927 Differentiation and Integration questions If it's $$\int \frac{\sqrt{\ln x}}{x} dx$$ ,then it's an easy one. U know that (i hope ) $$d(\ln x)=\frac{dx}{x}$$ It follows immediately that your integral can be written $$\int \sqrt{\ln x} d(\ln x)$$ ,which can be easily computed making the substitution $$\ln x\rightarrow u$$ Daniel.
 Sci Advisor HW Helper P: 2,002 1. What do you know about the expression $\cos^2x+\sin^2x$? (Some kind of identity maybe?) 2. Just integrate. Post what you have so far. Notice that $\frac{1}{(x+3)^3}=(x+3)^{-3}$ 3. What's the derivative of $(\ln x)^n$ for any n?. EDIT: Darn, beaten by 3 posts.
 Sci Advisor HW Helper P: 11,927 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": $$u(x)=:x+3$$ Then: $$u(2)=:2+3=5;u(5)=:5+3=8$$ 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: $$y(x)\rightarrow y(u(x))$$ 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.