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Recitation 10

Problem 1 Find the following indefinite integrals: (a) ∫x1−x2dx\int \frac{x}{\sqrt{1-x^2}}dx; (b) ∫x1−x2dx\int x\sqrt{1-x^2}dx; (c) ∫x21−x2dx\int x^2\sqrt{1-x^2}dx.

Solution (a) Let y=1−x2y = 1 - x^2. Thus dy=−2xdxdy = -2xdx. The integral is equal to ∫−12y−1/2dy=−y1/2+C=−1−x2+C.\int -\frac{1}{2}y^{-1/2}dy = -y^{1/2} + C = -\sqrt{1-x^2} + C. (b) The integral is equal to ∫−12y1/2dy=−13y3/2+C=−13(1−x2)3/2+C.\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⁡ux = \sin u and dx=cos⁡ududx = \cos u du. Then 1−x2=1−sin⁡2u=cos⁡u\sqrt{1-x^2} = \sqrt{1-\sin^2 u} = \cos u. The integral is equal to ∫sin⁡2ucos⁡2udu=14∫sin⁡22udu=18∫(1−cos⁡4u)du=18(u−14sin⁡4u)+C.\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⁡xu = \arcsin x and 14sin⁡4u=12sin⁡2ucos⁡2u=sin⁡ucos⁡u(1−2sin2u)=x1−x2(1−2x2).\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 18(arcsin⁡x−x1−x2(1−2x2))+C.\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) ∫2x+7(x−1)(x2−5x+6)dx\int \frac{2x+7}{(x-1)(x^2-5x+6)}dx; (b) ∫x3+x2−2x−4x2−5x+6dx\int \frac{x^3+x^2-2x-4}{x^2-5x+6}dx.

Solution (a) The partial fraction decomposition of 2x+7(x−1)(x2−5x+6)=Ax−1+Bx−2+Cx−3.\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−1x - 1 and plugging in x=1x = 1 gives A=92A = \tfrac{9}{2}. Similarly, we get B=−11,C=132B = -11, C = \tfrac{13}{2}. The integral is thus 92ln⁡∣x−1∣−11ln⁡∣x−2∣+132ln⁡∣x−3∣+C.\tfrac{9}{2}\ln |x-1| - 11 \ln|x-2| + \tfrac{13}{2} \ln |x-3| + C. (b) By long polynomial division, x3+x2−2x−4x2−5x+6=x+6+22x−40x2−5x+6\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 22x−40x2−5x+6=Ax−2+Bx−3.\frac{22x - 40}{x^2-5x+6} = \frac{A}{x-2} + \frac{B}{x-3}. Solve for A,BA, B and get A=−4,B=26A = -4, B = 26. The integral is thus 12x2+6x−4ln⁡∣x−2∣+26ln⁡∣x−3∣+C.\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): m∫cos⁡mxdx=cos⁡m−1xsin⁡x+(m−1)∫cos⁡m−2xdx;m∫sin⁡mxdx=−sin⁡m−1cos⁡x+(m−1)∫sin⁡m−2xdx.\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: In=∫cos⁡nxdx.I_n = \int \cos^n x dx. Now re-write as: In=∫cos⁡n−1xcos⁡xdx.I_n = \int \cos^{n-1}x \cos x dx. Integrating by this substitution: cos⁡xdx=d(sin⁡x),In=∫cos⁡n−1xd(sin⁡x)\cos x dx = d(\sin x), I_n = \int \cos^{n-1}x d(\sin x). Now integrating by parts: ∫cos⁡nxdx=cos⁡n−1xsin⁡x−∫sin⁡xd(cos⁡n−1x)=cos⁡n−1xsin⁡x+(n−1)∫sin⁡xcos⁡n−2xsin⁡xdx=cos⁡n−1xsin⁡x+(n−1)∫cos⁡n−2x(1−cos⁡2x)dx=cos⁡n−1xsin⁡x+(n−1)In−2−(n−1)In.\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 InI_n: nIn=cos⁡n−1xsin⁡x+(n−1)In−2nI_n = \cos^{n-1}x\sin x + (n-1)I_{n-2}.

Problem 3 Find the following indefinite integrals: (a) ∫x3sin⁡xdx\int x^3\sin x dx; (b) ∫cos⁡3xdx\int \cos^3 x dx; (c) ∫cos⁡4xdx\int \cos^4 x dx; (d) ∫1sin⁡2xdx\int \frac{1}{\sin^2 x} dx; (e) ∫1sin⁡xdx\int \frac{1}{\sin x} dx.

Solution (a) Using integration by parts, we obtain ∫x3sin⁡xdx=∫−x3dcos⁡x=−x3cos⁡x−∫cos⁡xd(−x3)=−x3cos⁡x+3∫x2cos⁡xdx.\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, ∫x2cos⁡xdx=∫x2dsin⁡x=x2sin⁡x−∫sin⁡xd(x2)=x2sin⁡x−2∫xsin⁡xdx.\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 ∫xsin⁡xdx=∫−xdcos⁡x=−xcos⁡x−∫cos⁡xd(−x)=−xcos⁡x+sin⁡x+C.\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(x2−2)sin⁡x−x(x2−6)cos⁡x+C.3(x^2-2)\sin x - x(x^2-6)\cos x + C. (b) By the reduction formula for m=3m = 3, the integral equals 13cos⁡2xsin⁡x+23∫cos⁡xdx=13cos⁡2xsin⁡x+23sin⁡x+C.\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=4m = 4, the integral equals 14cos⁡3xsin⁡x+34∫cos⁡2xdx.\frac{1}{4}\cos^3 x\sin x + \frac{3}{4}\int \cos^2 x dx. By the reduction formula for m=2m = 2, ∫cos⁡2xdx=12(cos⁡xsin⁡x+12∫dx)=12cos⁡xsin⁡x+12x+C.\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 14cos⁡3xsin⁡x+18cos⁡xsin⁡x+18x+C.\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=0m = 0, ∫sin−2xdx=−sin⁡−1cos⁡x+C=−cot⁡x+C.\int sin^{-2}x dx = -\sin^{-1}\cos x + C = -\cot x + C. (e) Substitute x=2ux = 2u and dx=2dudx = 2 du and get ∫1sin⁡2u2du=∫1sin⁡ucos⁡udu=∫sin⁡2u+cos⁡2usin⁡ucos⁡udu=∫tan⁡udu+∫cot⁡udu=−ln⁡∣cos⁡u∣+ln⁡∣sin⁡u∣+C=ln⁡∣tan⁡(x/2)∣+C.\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 ∫exsin⁡xdx=12ex(sin⁡x−cos⁡x)+C,∫excos⁡xdx=12ex(sin⁡x+cos⁡x).\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) ∫xexsin⁡xdx\int xe^x\sin x dx; (b) ∫x2exsin⁡xdx\int x^2e^{x}\sin x dx; (c) ∫ln⁡xx2dx\int \frac{\ln x}{x^2} dx.

Solution outline (a) Let f(x)=x,g(x)=12ex(sin⁡x−cos⁡x)f(x) = x, g(x) = \frac{1}{2}e^x(\sin x - \cos x). The integral equals ∫fdg=fg−∫gdf=12xex(sin⁡x−cos⁡x)−12∫ex(sin⁡x−cos⁡x)dx.\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)=x2,g(x)=12ex(sin⁡x−cos⁡x)f(x) = x^2, g(x) = \frac{1}{2}e^x(\sin x - \cos x). The integral equals ∫fdg=fg−∫gdf=12x2ex(sin⁡x−cos⁡x)−∫xex(sin⁡x−cos⁡x)dx.\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 ∫xexsin⁡xdx\int xe^x\sin x dx is given by part (a). The answer to ∫xexcos⁡xdx\int xe^x\cos x dx can be obtained similarly. (c) Integration by parts gives ∫ln⁡xx2dx=∫ln⁡xd(−1/x)=−ln⁡xx−∫(−1/x)d(ln⁡x)=−ln⁡xx+∫1/x2dx=−ln⁡xx−1x+C.\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 ⁣:[0,2]→Rf\colon [0,2] \to \mathbb{R} be the function defined by f(x)=0f(x)=0 for x∈[0,1]x \in [0,1] and f(x)=1f(x)=1 for x∈(1,2]x \in (1,2]. Show that ff does not have antiderivative.

Solution Assume, for the sake of contradiction, that ff has an antiderivative FF. For every 0<h<10 < h < 1, by mean value theorem, F(1+h)−F(1)h=F′(1+c)=f(1+c)\frac{F(1 + h) - F(1)}{h} = F'(1+c) = f(1+c) for some 0<c<h0 < c < h. Therefore F(1+h)−F(1)h=1\frac{F(1 + h) - F(1)}{h} = 1 for every 0<h<10 < h < 1, and so 0=f(1)=F′(1)=lim⁡h→0+F(1+h)−F(1)h=1.0 = f(1) = F'(1) = \lim_{h\to 0^+}\frac{F(1 + h) - F(1)}{h} = 1. A contradiction.

Problem 6 Let f ⁣:[0,2]→Rf\colon [0,2] \to \mathbb{R} be a function such that for all x∈[0,1]x \in [0,1] we have f(x)<0f(x)<0 and for all x∈(1,2]x \in (1,2] we have f(x)>0f(x)>0. Show that ff does not have antiderivative.

Solution Similar to problem 5, using mean value theorem, if the antiderivative exists, then F(1+h)−F(1)h>0\frac{F(1+h)-F(1)}{h} > 0 for all 0<h<10 < h < 1. Therefore 0>f(1)=F′(1)=lim⁡h→0+F(1+h)−F(1)h≥0.0 > f(1) = F'(1) = \lim_{h\to 0^+}\frac{F(1 + h) - F(1)}{h} \ge 0. A contradiction.