calculus
I am preparing for a basic level calculus test and came across this problem:
$$ \int_0^{\pi/2} \dfrac{1}{2-\cos{x}} dx$$
Which appears to be simple enough before I realize that I can't multiply the top and bottom by $ 2+\cos{x} $ and get anywhere fast. The prompt and Wolfram Alpha suggest a u-substitution of $ u = \tan{\frac{x}{2}} $ but I really have no idea how to implement that. How is this substitution used? Is this even an appropriate problem for my current level of math knowledge?
If you are looking for some challenging calculus problems, I would suggest looking at the HMMT (Harvard-MIT Mathematics Tournament) problems archive. Take a look at the calculus test for the years 1998-2011. I don't think these problems are the routine exercises that you have grown weary of in Thomas' Calculus.
If you want another book that may suit your taste, I recommend "Introduction to Calculus and Analysis" by Richard Courant and Fritz John. I think the problems are a bit harder than those in Thomas' text. But more importantly, Courant and John give a lot of motivation for calculus through discussing its relevance to the physical world (there is even discussion of Fourier Series in the study of a vibrating drum).
Courant and John also introduce some concepts of modern analysis (calculus done more rigorously) such as monotone sequences, a precise discussion of a limit, etc towards the end of the book, if that interests you.
Last but not least, I see that you mentioned the Putnam. I think a great book for doing calculus problems that may prepare you for the Putnam is Titu Andreescu's book- "Problems in Real Analysis: Advanced Calculus on the Real Line".
Andreescu's texts are known to be useful in preparing students for mathematics competitions. In this textbook, Andreescu exposes you to a lot of techniques (especially in dealing with inequalities) that may help you with the Putnam. You may also learn some analysis along the way, but since he is dealing only with the real line, he doesn't "weigh you down" with various topological concepts that may or may not interest you. If you have mastered the concepts of calculus and are looking for a challenge, this book may be for you.
Good luck!
Refer to the figure:
The volume of the container filled with oil is: $$V=\frac{(0.3+2x)+0.3}{2}\cdot h\cdot 10.$$
From the similarity of the two triangles on the left we get: $$\frac{x}{0.25}=\frac{h}{0.5} \Rightarrow x=\frac{h}{2}.$$
Substitute this into the volume formula: $$V=\frac{(0.3+2\cdot \frac{h}{2})+0.3}{2}\cdot h\cdot 10=\frac{(h+0.6)h}{2}\cdot 10=5h^2+3h.$$ Take the derivative of the volume function with respect to time: $$\frac{dV}{dt}=\frac{dV}{dh}\cdot \frac{dh}{dt}=2 \ \frac{m^3}{\text{min}} \Rightarrow \\ (10h+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ (10\cdot 0.3+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ \frac{dh}{dt}=\frac{2}{6}=\frac13.$$
Best Answer
hint: use the sub $u = \tan( x/2), \cos t = \frac{1-u^2}{1+u^2}$ and some partial fraction.
differencing you get $$du = (1 + u^2) dx/2, x = 0 \to u = 0, x = \pi/2\to u = 1$$ therefore the $\int_0^{\pi/2} \frac{dx}{2-\cos x}$ is transformed into $$\int_0^1 \frac{2du}{(1 + u^2} \frac{1}{\left(2 - \frac{1-u^2}{1+u^2}\right)} =2 \int_0^1\frac{du}{3u^2 + 1} = \frac{2}{3} \int_0^1 \frac{du}{u^2 + 1/3}$$ can you continue now?