**Problem 1** Prove the reduction formula for n\neq 1: \int\tan^n x dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2} x dx.

**Idea** Rewrite \tan^{n}x = \tan^{n-2}x(\sec^2 x - 1) and use (\tan x)' = \sec^2x.

**Fact 1 **If g(t) = \int_{a(t)}^{b(t)} f(x) dx, then g'(t) = f(a(t))a'(t) - f(b(t))b'(t). Prove it by the fundamental theorem of calculus.

**Problem 2** Consider the function f\colon [0,1] \to \mathbb{R} defined by f(x) = e^{x^2}. For every 0 < t < 1 let A(t) denote the area below the graph of f and above the x-axis for 0<x<t. Let B(t)be the area above the graph of f and below the line y = e for t < x < 1. Find the maximum point of A(t)+B(t).

**Solution **Note that A(t) = \int_0^t f(x) dx and B(t) = \int_t^1 (e - f(x)) dx. Let S(t) = A(t) + B(t). By the fundamental theorem of calculus, S'(t) = A'(t) + B'(t) = f(t) - (e - f(t)) = 2f(t) - e. Solve S'(t) = 0 and get t^* = \sqrt{1-\ln 2}. When 0 \le t < t^*, S'(t) < 0 and so S is monotone decreasing; when t^* < t \le 1, S'(t) > 0 and so S is monotone increasing. Therefore t = 0, 1 are local maximum points. We only need to compare S(0) and S(1). Because S(0) + S(1) = e, it is enough to compare S(1) with e/2. Notice that S(1) = \int_0^1 f(x) dx and we can approximate the integral from below by (f(0) + f(1/8) + f(2/8) + f(3/8) + f(4/8) + f(5/8) + f(6/8) + f(7/8))/8 = 1.36... > e/2. Therefore S(1) > e/2 > S(0) and t=1 is the maximum point.

**Alternative solution** The last part to show \int_0^1 e^{x^2}dx > e/2 can be simplified as follows. Because 0 \le x \le 1 and e^{x^2} \ge 1, e^{x^2} - 1 \ge (e^{x^2} - 1)x and so e^{x^2} \ge \left(e^{x^2} - 1\right)x + 1. In other words, the graph of e^{x^2} is above the graph of \left(e^{x^2} - 1\right)x + 1 for 0 \le x \le 1. Therefore \int_0^1 e^{x^2} dx > \int_0^1 \left[\left(e^{x^2} - 1\right)x + 1\right] dx = \left.\frac{1}{2}e^{x^2} - \frac{1}{2}x^2 + x \right|_0^1 = e/2.

**Problem 3** Find the derivative of F(x) = \int_{0}^{x^2}|\sin t|dt.

**Answer **F'(x) = |\sin x^2| 2x.

**Problem 4** Find the limit \lim_{x\to 0}\frac{\int_0^x (1-\cos t)dt}{x^2\sin x}dt.

**Answer** We know that \frac{\int_0^x (1-\cos t)dt}{x^2\sin x} = \frac{\int_0^x (1-\cos t)dt}{x^3} \frac{x}{\sin x}. It suffices to compute \lim_{x\to 0}\frac{\int_0^x (1-\cos t)dt}{x^3}. This limit is of type 0/0. We consider \lim_{x\to 0}\frac{\left(\int_0^x (1-\cos t)dt\right)'}{\left(x^3\right)'} = \lim_{x\to 0}\frac{1-\cos x}{3x^2}, which is again a limit of type 0/0. We consider \lim_{x\to 0}\frac{\left(1-\cos x\right)'}{\left(3x^2\right)'} = \lim_{x\to 0}\frac{\sin x}{6x} = \frac{1}{6}. By l’Hôpital’s rule twice, we get the limit is 1/6.

**Problem 5 **Consider the one-to-one and onto function f\colon [0,2] \to [1,6] defined by f(x) = x + 2^x. Show that \int_0^2 f(x) dx + \int_1^6 f^{-1}(x) dx = 12.

**Geometric solution** Observe that \int_0^2 f(x) dx is the area bounded by y = f(x), y = 0 and x = 2, and \int_1^6 f^{-1}(x) dx is the area bounded by y = f(x), x = 0 and y = 6. Thus the sum is the area of the rectangle bounded by x = 0, 2 and y = 1,6.

**Algebraic solution** For the second integral \int_1^6 f^{-1}(x) dx consider substitution y = f^{-1}(x). Thus x = f(y), dx = f'(y) dy and \int_1^6 f^{-1}(x) dx = \int_0^2 yf'(y)dy. Using integration by parts, \int_0^2 yf'(y)dy = \left.yf(y)\right|_0^2 - \int_0^2 f(y)dy. Thus \int_0^2 f(x) dx + \int_1^6 f^{-1}(x) dx = \int_0^2 f(y)dy + \int_0^2 yf'(y)dy = \left.yf(y)\right|_0^2 = 12.

**Problem 6 **Find the area bounded by the arcs of the parabolas y = x^2 / 4 and y = 3 - x^2 / 2.

**Idea** The intersections of the parabolas are given by x^2 / 4 = 3 - x^2 / 2, that is x = \pm 2. The area is thus \int_{-2}^2 3 - x^2 /2 - x^2 / 4.

**Problem 7 **Compute the integral \int_4^9 \frac{\sqrt{x}}{\sqrt{x} - 1}dx.

**Idea **The anti-derivative of \frac{\sqrt{x}}{\sqrt{x} - 1} is x + 2\sqrt{x} + 2\ln(\sqrt{x}-1). By Newton–Leibniz formula, the definite integral equal \left.x + 2\sqrt{x} + 2\ln(\sqrt{x}-1)\right|_4^9 = 7+\ln 4.

**Problem 8** Find the length of the graph of f(x) = x^2 for 0\le x \le 1.

**Solution **The length is given by \int_0^1 \sqrt{1 + f'(x)^2} dx = \int_0^1 \sqrt{1 + 4x^2} dx. The integral of \int_0^1 \sqrt{1 + 4x^2} dx can be obtained by Euler substitution \sqrt{1 + 4x^2} = t + 2x. We get x = \frac{1-t^2}{4t}, t+2x = \frac{1+t^2}{2t} and dx = \frac{1+t^2}{4t^2}dt. Thus \int \sqrt{1 + 4x^2} dx = \int \frac{(1+t^2)^2}{8t^3}dt = \frac{t^4 + 4t^2\ln t - 1}{16t^2}. On the other hand, by our substitution t = \sqrt{1 + 4x^2} - 2x. Plug this in and get \int \sqrt{1 + 4x^2} dx = \frac{1}{2}x\sqrt{1+4x^2} + \frac{1}{4}\ln (\sqrt{1 + 4x^2} - 2x). By Newton–Leibniz formula, the definite integral equals \frac{1}{2}\sqrt{5} + \frac{1}{4}\ln (\sqrt{5}-2).

**Alternative solution** We can also use trigonometric substitution x = \frac{1}{2}\tan u for \int \sqrt{1 + 4x^2} dx and use reduction formula for \int 1/\cos^3 u du.

**Problem 9** Prove that for any function f\colon [0,1]\to \mathbb{R} \int_0^\pi xf(\sin x) dx = \frac{\pi}{2}\int_0^\pi f(\sin x)dx.

**Algebraic solution **Use substitution x = \pi - u and get \int_0^\pi xf(\sin x) dx = \int_\pi^0 (\pi - u)f(\sin u)(-du) = \int_0^\pi (\pi - u)f(\sin u)du. Therefore 2\int_0^\pi xf(\sin x) dx = \int_0^\pi xf(\sin x) dx + \int_0^\pi (\pi - u)f(\sin u)du = \int_0^\pi \pi f(\sin x) dx.

**Question **Can you interpret the algebraic solution geometrically?

**Problem 10** Let A be the area bounded between the graph of f (x) = \sin x and the graph of g(x)=\cos x where \pi/4 \le x \le 2\pi +\pi /4. Find the volume of the solid obtained by revolving A about the horizontal line y = 4.

**Idea **Notice that the graphs of \sin x is above the graph of \cos x for \pi / 4 \le x \le \pi + \pi/4 and the other way around for \pi+\pi / 4 \le x \le 2\pi + \pi/4. The volume is thus given by \int_{\pi/4}^{\pi + \pi / 4} \pi\left((4 - \cos x)^2 - (4 - \sin x)^2\right) dx + \int_{\pi + \pi/4}^{2\pi + \pi / 4} \pi\left((4 - \sin x)^2 - (4 - \cos x)^2\right) dx.