Problem 1 Find the following indefinite integrals: (a) \int \frac{x}{\sqrt{1-x^2}}dx; (b) \int x\sqrt{1-x^2}dx; (c) \int x^2\sqrt{1-x^2}dx.
Solution (a) Let y = 1 - x^2. Thus dy = -2xdx. The integral is equal to \int -\frac{1}{2}y^{-1/2}dy = -y^{1/2} + C = -\sqrt{1-x^2} + C. (b) The integral is equal to \int -\frac{1}{2}y^{1/2}dy = -\frac{1}{3}y^{3/2} + C = -\frac{1}{3}(1-x^2)^{3/2}+C. (c) Substitute x = \sin u and dx = \cos u du. Then \sqrt{1-x^2} = \sqrt{1-\sin^2 u} = \cos u. The integral is equal to \int \sin^2u\cos^2u du = \frac{1}{4}\int \sin^2 2u du = \frac{1}{8}\int (1-\cos 4u)du = \frac{1}{8}(u - \tfrac{1}{4}\sin 4u) + C. Note that u = \arcsin x and \tfrac{1}{4}\sin 4u = \tfrac{1}{2}\sin 2u\cos 2u = \sin u\cos u(1-2sin^2 u) = x\sqrt{1-x^2}(1-2x^2). Therefore the integral is equal to \frac{1}{8}\left(\arcsin x - x\sqrt{1-x^2}\left(1-2x^2\right)\right) + C.
Problem 2 Find the following indefinite integrals: (a) \int \frac{2x+7}{(x-1)(x^2-5x+6)}dx; (b) \int \frac{x^3+x^2-2x-4}{x^2-5x+6}dx.
Solution (a) The partial fraction decomposition of \frac{2x+7}{(x-1)(x^2-5x+6)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}. Multiplying both sides by x - 1 and plugging in x = 1 gives A = \tfrac{9}{2}. Similarly, we get B = -11, C = \tfrac{13}{2}. The integral is thus \tfrac{9}{2}\ln |x-1| - 11 \ln|x-2| + \tfrac{13}{2} \ln |x-3| + C. (b) By long polynomial division, \frac{x^3+x^2-2x-4}{x^2-5x+6} = x + 6 + \frac{22x - 40}{x^2-5x+6}. The partial fraction decomposition of \frac{22x - 40}{x^2-5x+6} = \frac{A}{x-2} + \frac{B}{x-3}. Solve for A, B and get A = -4, B = 26. The integral is thus \tfrac{1}{2}x^2 + 6x - 4\ln |x-2| + 26\ln |x-3| + C.
Reduction formulas For indefinite integrals of powers of trig functions, we have two useful reduction formulas (try to prove them by integration by parts): \begin{aligned}m\int \cos^m x dx &= \cos^{m-1}x\sin x + (m-1)\int \cos^{m-2}x dx; \\ m\int \sin^m x dx &= -\sin^{m-1}\cos x + (m-1)\int \sin^{m-2}x dx.\end{aligned}
Proof of the 1st reduction formula Start by setting: I_n = \int \cos^n x dx. Now re-write as: I_n = \int \cos^{n-1}x \cos x dx. Integrating by this substitution: \cos x dx = d(\sin x), I_n = \int \cos^{n-1}x d(\sin x). Now integrating by parts: \begin{aligned}\int \cos^n x dx & = \cos^{n-1}x \sin x - \int \sin x d(\cos^{n-1}x) \\ &= \cos^{n-1}x \sin x + (n-1)\int \sin x \cos^{n-2} x \sin x dx \\ & = \cos^{n-1}x\sin x + (n-1)\int \cos^{n-2} x(1-\cos^2 x)dx \\ &= \cos^{n-1}x\sin x + (n-1)I_{n-2} - (n-1)I_n. \end{aligned} Solving for I_n: nI_n = \cos^{n-1}x\sin x + (n-1)I_{n-2}.
Problem 3 Find the following indefinite integrals: (a) \int x^3\sin x dx; (b) \int \cos^3 x dx; (c) \int \cos^4 x dx; (d) \int \frac{1}{\sin^2 x} dx; (e) \int \frac{1}{\sin x} dx.
Solution (a) Using integration by parts, we obtain \int x^3\sin x dx = \int -x^3d\cos x = -x^3\cos x - \int \cos x d(-x^3) = -x^3\cos x + 3\int x^2\cos x dx. By integration by parts again, \int x^2\cos x dx = \int x^2 d\sin x = x^2\sin x - \int \sin x d(x^2) = x^2\sin x - 2\int x\sin x dx. Again, integration by parts gives \int x\sin x dx = \int -xd\cos x = -x\cos x - \int \cos xd(-x) = -x\cos x + \sin x + C. The integral is thus 3(x^2-2)\sin x - x(x^2-6)\cos x + C. (b) By the reduction formula for m = 3, the integral equals \frac{1}{3}\cos^2 x\sin x + \frac{2}{3}\int \cos x dx = \frac{1}{3}\cos^2 x\sin x + \frac{2}{3}\sin x + C. (c) By the reduction formula for m = 4, the integral equals \frac{1}{4}\cos^3 x\sin x + \frac{3}{4}\int \cos^2 x dx. By the reduction formula for m = 2, \int \cos^2 x dx = \frac{1}{2}\left(\cos x \sin x + \frac{1}{2}\int dx\right) = \frac{1}{2}\cos x\sin x + \frac{1}{2}x + C. The integral is thus \frac{1}{4}\cos^3 x \sin x + \frac{1}{8} \cos x \sin x + \frac{1}{8} x + C. (d) By the reduction formula for m = 0, \int sin^{-2}x dx = -\sin^{-1}\cos x + C = -\cot x + C. (e) Substitute x = 2u and dx = 2 du and get \begin{aligned}\int \frac{1}{\sin 2u} 2du &= \int \frac{1}{\sin u\cos u}du \\ &= \int \frac{\sin^2 u + \cos^2 u}{\sin u\cos u}du \\ &= \int \tan u du + \int \cot u du \\ &= -\ln |\cos u| + \ln |\sin u| + C \\ &= \ln |\tan (x/2)| + C.\end{aligned}
Fact Using integration by parts, we can get \int e^x\sin x dx = \frac{1}{2}e^x(\sin x - \cos x) + C, \int e^x\cos x dx = \frac{1}{2}e^x(\sin x + \cos x).
Problem 4 Find the following indefinite integrals: (a) \int xe^x\sin x dx; (b) \int x^2e^{x}\sin x dx; (c) \int \frac{\ln x}{x^2} dx.
Solution outline (a) Let f(x) = x, g(x) = \frac{1}{2}e^x(\sin x - \cos x). The integral equals \int f dg = fg - \int g df = \frac{1}{2}xe^x(\sin x - \cos x) - \frac{1}{2}\int e^x(\sin x - \cos x) dx. Use the fact twice then. (b) Let f(x) = x^2, g(x) = \frac{1}{2}e^x(\sin x - \cos x). The integral equals \int f dg = fg - \int g df = \frac{1}{2}x^2e^x(\sin x - \cos x) - \int xe^x(\sin x - \cos x)dx. The answer to \int xe^x\sin x dx is given by part (a). The answer to \int xe^x\cos x dx can be obtained similarly. (c) Integration by parts gives \int \frac{\ln x}{x^2} dx = \int \ln x d(-1/x) = -\frac{\ln x}{x} - \int (-1/x)d(\ln x) = -\frac{\ln x}{x} + \int 1/x^2dx = -\frac{\ln x}{x} - \frac{1}{x} + C.
Problem 5 Let f\colon [0,2] \to \mathbb{R} be the function defined by f(x)=0 for x \in [0,1] and f(x)=1 for x \in (1,2]. Show that f does not have antiderivative.
Solution Assume, for the sake of contradiction, that f has an antiderivative F. For every 0 < h < 1, by mean value theorem, \frac{F(1 + h) - F(1)}{h} = F'(1+c) = f(1+c) for some 0 < c < h. Therefore \frac{F(1 + h) - F(1)}{h} = 1 for every 0 < h < 1, and so 0 = f(1) = F'(1) = \lim_{h\to 0^+}\frac{F(1 + h) - F(1)}{h} = 1. A contradiction.
Problem 6 Let f\colon [0,2] \to \mathbb{R} be a function such that for all x \in [0,1] we have f(x)<0 and for all x \in (1,2] we have f(x)>0. Show that f does not have antiderivative.
Solution Similar to problem 5, using mean value theorem, if the antiderivative exists, then \frac{F(1+h)-F(1)}{h} > 0 for all 0 < h < 1. Therefore 0 > f(1) = F'(1) = \lim_{h\to 0^+}\frac{F(1 + h) - F(1)}{h} \ge 0. A contradiction.