Chapter 2-B PDF - Indefinite Integrals and Substitution Method

## 5.5 Indefinite Integrals and the substitution Method. - $\int f(x)dx = F(x) + C$, where - $F'(x) = f(x)$ & C is Constant. ## Substitution: Running the Chain Rule Backwards. - $\frac{d}{dx}( \frac{u^{n+1}}{n+1}) = u^{n}. \frac{du}{dx}$, where $u = f(x)$ ## Integration - $\int \frac{d}{dx}...

## 5.5 Indefinite Integrals and the substitution Method. - $\int f(x)dx = F(x) + C$, where - $F'(x) = f(x)$ & C is Constant. ## Substitution: Running the Chain Rule Backwards. - $\frac{d}{dx}( \frac{u^{n+1}}{n+1}) = u^{n}. \frac{du}{dx}$, where $u = f(x)$ ## Integration - $\int \frac{d}{dx}(\frac{u^{n+1}}{n+1})dx = \int u^{n}. \frac{du}{dx}dx$ - $\implies \int u^{n}. \frac{du}{dx}dx = \frac{u^{n+1}}{n+1} + C$ ## Example 1: $\int (x^{2} + x)(3x + 1)dx$ - Solution: Put $u = x^{2} + x$, then $du = \frac{du}{dx}dx$ - $du = (3x + 1)dx$ - So by substitution we get: - $\int (x^{2} + x)(3x + 1)dx = \int udu = \frac{u^{6}}{6} + C$ - $= \frac{1}{6}(x^{2} + x)^{6} + C$ ## Example 2: $\int \sqrt{2x + 1} dx = \int (2x + 1)^{\frac{1}{2}} dx$ - Solution: Put $u = 2x + 1$, then $du = \frac{du}{dx}dx$ - $du = 2. dx$ - $\implies \int \sqrt{2x + 1} dx = \frac{1} {2}\int \sqrt{2x + 1}. 2dx$ - $=\frac{1}{2} \int u^{\frac{1}{2}} du$ - $= \frac{1}{2} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C$ - $= \frac{1}{3} u^{\frac{3}{2}} + C$ - $= (\frac{1}{3}(2x + 1)^{\frac{3}{2}} + C$ ## Theorem 6. The Substitution Rule - If $u = g(x) $ is a differentiable function, and - $f$ is continuous on $[a, b]$, - Then - $\int f(g(x)). g'(x)dx = \int f(u) du$. ## The substitution Method to evaluate $\int f(g(x))g'(x)dx$ 1. Substitution $u = g(x) $ and $du = \frac{du}{dx}dx$ - $\implies du = g'(x) dx$ 2. Integrate with respect to u to obtain $\int F(u)du$. 3. Replace u by g(x). ## Example 3: $\int sec^{2}(5x + 1). 5dx$ - Solution: 1. Substitution: $u = 5x + 1 \implies du = \frac{du}{dx}dx $ - $\implies du = 5dx$ 2. Integration wrt u: $\int sec^{2}udu = tanu + C$ 3. Replace u = 5x + 1 $\implies (tan(5x + 1) + C$ ## Example 4: $\int cos(7\theta + 3) d\theta$ - Solution: - $u = 7\theta + 3 \implies du = \frac{du}{d\theta}d\theta$ - $du = 7. d\theta$ - $\implies \int cos(7\theta + 3) d\theta = \frac{1}{7} \int cos(7\theta + 3). 7d\theta$ - $ = \frac{1}{7} \int cos u du$ - $= \frac{1}{7} sin u + C$ - $ = \frac{1}{7} sin (7\theta + 3) + C$ ## Example 5: $\int x^{2} cos x^{3} dx$ - Solution: - $u = x^{3} \implies du = \frac{du}{dx}.dx$ - $du = 3x^{2}. dx$ - $\implies \int x^{2} cos x^{3} dx = \frac{1}{3}\int cos x^{3}. 3x^{2}dx$ - $= \frac{1}{3} \int cos u du$ - $= \frac{1}{3} sin u + C$ - $= \frac{1}{3} sin x^{3} + C$ ## Example 6: $\int x \sqrt{2x + 1} dx$ - Solution: - $u = 2x + 1 \implies du = \frac{du}{dx}dx$ - $x = \frac{u-1}{2}$ - $du = 2 dx$ - $\implies \int x (2x + 1)^{\frac{1}{2}}. dx = \frac{1}{2}\int x(2x + 1)^{\frac{1}{2}}. 2dx$ - $ = \frac{1}{2} \int (\frac{u-1}{2}). u^{\frac{1}{2}} du$ - $ = \frac{1}{4} \int (u - 1). u^{\frac{1}{2}} du$ - $ = \frac{1}{4} \int (u^{\frac{3}{2}} - u^{\frac{1}{2}}) du$ - $ = \frac{1}{4} (\frac{u^{\frac{5}{2}}}{\frac{5}{2}} - \frac{u^{\frac{3}{2}}}{\frac{3}{2}}) + C$ - $ = \frac{1}{4} (\frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}}) + C$ - $ = \frac{1}{10}(2x + 1)^{\frac{5}{2}} - \frac{1}{6}(2x + 1)^{\frac{3}{2}} + C$ ## Example 7: - a) $\int sin^{2}x dx$ - Solution: - $\int sin^{2}x dx = \frac{1}{2} \int (1 - cos2x) dx$ - Note: $sin^{2}x = \frac{1}{2}(1 - cos2x)$ - $= \frac{1}{2} \int 1 dx - \frac{1}{2} \int cos2x dx$ - $ = \frac{1}{2}x - \frac{1}{4} sin 2x + C$ - b) $\int cos^{2}x dx$ - Solution: -$ = \frac{1}{2} \int (1+ cos2x) dx$ - Note: $cos^{2}x = \frac{1}{2}(1 + cos2x)$ - $ = \frac{1}{2}\int 1 dx + \frac{1}{2} \int cos2x dx$ - $ = \frac{1}{2}x + \frac{1}{4} sin 2x + C$ - c) $\int (1 - 2sinx) sin^{2}xdx$ - $ = \int (sin^{2}x + cos^{2}x - sin x) sin 2x dx$ - Note: $1 = sin^{2}x + cos^{2}x$ - $ = \int (cos^{2}x - sin x) sin 2x dx$ - Note: $cos 2x = cos^{2}x - sin^{2}x$ - $ = \int cos 2x sin2x dx$ - Note: $cos 2x sin2x = \frac{1}{2} sin 4x$ - $ = \int \frac{1}{2} sin 4x dx $ - $ = \frac{1}{2} . \frac{-1}{4} \int sin 4x. 4dx$ - $ = \frac{-1}{8} cos(4x) + C$ ## Example 8: $\int \frac{2z}{\sqrt{z^{2} + 1}} dz$ can be written by $\int 2z (z^{2} + 1)^{\frac{-1}{2}}dz $ - Solution: - Put $u = z^{2} + 1 \implies du = \frac{du}{dz}dz$ - $du = 2z dz$ - $\implies \int (z^{2} + 1)^{\frac{-1}{2}}. 2z dz$ - $ = \int u^{\frac{-1}{2}} du$ - $ = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C$ - $ = 2 u^{\frac{1}{2}} + C$ - $ = 2 (z^{2} + 1)^{\frac{1}{2}} + C$ - $ = 2 \sqrt{(z^{2} + 1)} + C$ ## Exercises: 5.5 - **1/245:** $\int 2(2x + 4)^{5} dx$, $u = 2x + 4$ - Sol: $u = 2x + 4 \implies du = 2 dx$ - $\implies \int (2x + 4)^{5}. 2dx = \int u^{5}. du = \frac{u^{6}}{6} + C = \frac{(2x + 4)^{6}}{6} + C$ - **3/245:** $\int 2x (x + 5)^{4} dx$, $u = x + 5$ - Sol: $u = x + 5 \implies du = 2x dx$ - $\implies \int (x + 5)^{4}. 2x dx = \int u^{4} du = \frac{u^{5}}{5} + C = \frac{(x + 5)^{5}}{5} + C$ - **5/245:** $\int (3x + 2)(3x^{2} + 4x)^{4} dx$, $u = 3x^{2} + 4x$ - Sol: $u = 3x^{2} + 4x \implies du = (6x + 4) dx \implies du = 2 (3x + 2)dx$ - $\implies \int (3x^{2} + 4x)^{4}. (3x + 2)dx = \frac{1}{2} \int (3x^{2} + 4x)^{4}. 2(3x + 2) dx$ - $= \frac{1}{2} \int u^{4} du$ - $= \frac{1}{2} \frac{u^{5}}{5} + C$ - $= \frac{1}{10} (3x^{2} + 4x)^{5} + C$ - **7/245:** $\int sin 3x dx$, $u = 3x$ - Sol. $u = 3x \implies du = 3 dx$ - $\int sin 3x dx = \frac{1}{3} \int sin 3x. 3 dx = \frac{1}{3} \int sinu du$ - $ = \frac{-1}{3} cosu + C = \frac{-1}{3} cos 3x + C$ - **9/245:** $\int sec(2t) tan(2t) dt$, $u = 2t$ - Sol: $u = 2t \implies du = 2 dt $ - $\int sec(2t) tan(2t) dt = \frac{1}{2} \int sec(2t) tan(2t). 2dt$ - $ = \frac{1}{2} \int secu. tanu. du$ - $ = \frac{1}{2} secu + C$ - $ = \frac{1}{2} sec 2t + C$ - **11/245:** $\int \frac{9x^{2}}{\sqrt{1 - x^{3}}} dx$, $u = 1 - x^{3}$ - Sol: $u = 1 - x^{3} \implies du = -3x^{2}dr$ - $\int \frac{9x^{2}}{\sqrt{1 - x^{3}}} dx = \int (1 - x^{3})^{\frac{-1}{2}}. (-3x^2)dx = -3 \int u^{\frac{-1}{2}} du$ - $ = -3 \int u^{\frac{-1}{2}}du = -3( \frac{u^{\frac{1}{2}}}{\frac{1}{2}}) + C$ - $ = -6 \sqrt{u} + C$ - $ = -6\sqrt{1 - x^{3}} +C$ - **17/245:** $\int \sqrt{3 - 2S} dS$ - Sol: Put $u = 3 - 2S \implies du = -2 dS$ - $\int \sqrt{3 - 2S} dS = \frac{-1}{2} \int (3 - 2S)^{\frac{1}{2}} . (-2) dS$ - $ = \frac{-1}{2} \int u^{\frac{1}{2}}du$ - $ = \frac{-1}{2}(\frac{u^{\frac{3}{2}}}{\frac{3}{2}}) + C$ - $ = \frac{-1}{3}(3-2S)^{\frac{3}{2}} + C$ - **19/245:** $\int \theta \sqrt[4]{1 - \theta^{2}} d\theta = \int \theta (1 - \theta^{2})^{\frac{1}{4}} d\theta$ - Sol: Put $u = 1 - \theta^{2} \implies du = -2\theta d\theta $ - $\int (1 - \theta^{2})^{\frac{1}{4}} d\theta = \frac{-1}{2} \int (1 - \theta^{2})^{\frac{1}{4}}. (-2\theta d\theta)$ - $ = \frac{-1}{2} \int u^{\frac{1}{4}} du = \frac{-1}{2}(\frac{u^{\frac{5}{4}}}{\frac{5}{4}}) + C$ - $ = \frac{-2}{5}(1 - \theta^{2})^{\frac{5}{4}} + C$ - **23/245:** $\int sec^{2}(3x +2) dx$ - Put $u = 3x + 2 \implies du = 3 dx$ - $\int sec^{2}(3x +2) dx = \frac{1}{3} \int sec^{2}(3x+2). 3 dx$ - $= \frac{1}{3}\int secu du$ - $= \frac{1}{3} tan u + C$ - $= \frac{1}{3} tan (3x + 2) + C$ - **37/245:** $\int \frac{x}{\sqrt{1 + x}} dx$ - Put $u = 1 + x \implies du = 1. dx$ - $\int \frac{x}{\sqrt{1 + x}} dx = \int \frac{u - 1}{\sqrt{u}}. \frac{dx}{\frac{du}{dx}} du$ - $ = \int \frac{(u - 1)}{\sqrt{u}} du = \int u^\frac{-1}{2}(u - 1) du$ - $ = \int (u^\frac{1}{2} - u^\frac{-1}{2}) du$ - $ = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} - \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C$ - $ = \frac{2}{3}(1 + x)^{\frac{3}{2}} - 2 (1 + x)^{\frac{1}{2}} + C$ - **49/245:** $\int \frac{x}{(x^{2} - 4)^{3}} dx$ - Put $u = x^{2} - 4 \implies du = 2x dx$ - $\int \frac{x}{(x^{2} - 4)^{3}} dx = \frac{1}{2} \int (x^{2} - 4)^{-3}. 2x dx = \frac{1}{2} \int u^{-3} du$ - $ = \frac{1}{2} \frac{u^{-2}}{-2} + C$ - $ = \frac{-1}{4} (x^{2} - 4)^{-2} + C = \frac{-1}{4(x^{2} - 4)^{2}} + C$