Review of Integration Techniques

Example. Compute $\displaystyle \int \dfrac{7x^3 - 2x^2 + 12 - 8}{x^2(x^2 + 4)}\,dx$ .

Use partial fractions:

$$\dfrac{7x^3 - 2x^2 + 12 - 8}{x^2(x^2 + 4)} = \dfrac{ax + b}{x^2 + 4} + \dfrac{c}{x} + \dfrac{d}{x^2},$$

$$7x^3 - 2x^2 + 12x - 8 = (ax + b)x^2 + cx(x^2 + 4) + d(x^2 + 4).$$

Let $x = 0$ . Then $-8 = 4d$ , so $d = -2$ . Therefore,

$$7x^3 - 2x^2 + 12x - 8 = (ax + b)x^2 + cx(x^2 + 4) - 2(x^2 + 4).$$

Differentiate:

$$21x^2 - 4x + 12 = (ax + b)(2x) + ax^2 + cx(2x) + c(x^2 + 4) - 4x.$$

Let $x = 0$ . Then $12 = 4c$ , so $c = 3$ . Therefore,

$$21x^2 - 4x + 12 = (ax + b)(2x) + ax^2 + 3x(2x) + 3(x^2 + 4) - 4x,$$

$$21x^2 - 4x + 12 = (ax + b)(2x) + ax^2 + 6x^2 + 3(x^2 + 4) - 4x.$$

Differentiate:

$$42x - 4 = (ax + b)(2) + a(2x) + a(2x) + 12x + 6x - 4.$$

Let $x = 0$ . Then $-4 = 2b - 4$ , so $b = 0$ . Therefore,

$$42x - 4 = (ax)(2) + a(2x) + a(2x) + 12x + 6x - 4.$$

Let $x = 1$ . Then $38 = 6a + 14$ , so $a = 4$ .

Therefore,

$$ \int \dfrac{7x^3 - 2x^2 + 12 - 8}{x^2(x^2 + 4)}\,dx = \int \left(\dfrac{4x}{x^2 + 4} + \dfrac{3}{x} - \dfrac{2}{x^2}\right)\,dx = 2\ln |x^2 + 4| + 3\ln |x| + \dfrac{2}{x} + C.\quad\halmos$$

Example. Compute $\displaystyle \int (\cos 3x)^4\,dx$ .

$$\int (\cos 3x)^4\,dx = \int \left((\cos 3x)^2\right)^2\,dx = \int \left(\dfrac{1}{2}(1 + \cos 6x)\right)^2\,dx = \dfrac{1}{4} \int \left(1 + 2\cos 6x + (\cos 6x)^2\right)\,dx =$$

$$\dfrac{1}{4} \int \left(1 + 2\cos 6x + \dfrac{1}{2}(1 + \cos 12x)\right)\,dx = \dfrac{1}{4} \int \left(\dfrac{3}{2} + 2\cos 6x + \dfrac{1}{2}\cos 12x\right)\,dx =$$

$$\dfrac{3}{8}x + \dfrac{1}{12}\sin 6x + \dfrac{1}{96}\sin 12x + C. \quad\halmos$$

Example. Compute $\displaystyle \int (\cos 3x)^4(\sin 3x)^3\,dx$ .

$$\int (\cos 3x)^4(\sin 3x)^3\,dx = \int (\cos 3x)^4(\sin 3x)^2(\sin 3x\,dx) = \int (\cos 3x)^4\left(1 - (\cos 3x)^2\right)(\sin 3x\,dx) =$$

$$\left[u = \cos 3x, \quad du = -3\sin 3x\,dx, \quad dx = \dfrac{du}{-3\sin 3x}\right]$$

$$\int u^4(1 - u^2)(\sin 3x)\left(\dfrac{du}{-3\sin 3x}\right) = \dfrac{1}{3} \int (u^6 - u^4)\,du = \dfrac{1}{3}\left(\dfrac{1}{7}u^7 - \dfrac{1}{5}u^5\right) + C =$$

$$\dfrac{1}{3}\left(\dfrac{1}{7}(\cos 3x)^7 - \dfrac{1}{5}(\cos 3x)^5\right) + C.\quad\halmos$$

Example. Compute $\displaystyle \int \dfrac{\sqrt{1 + x^2}}{x^4}\,dx$ .

$$\int \dfrac{\sqrt{1 + x^2}}{x^4}\,dx = \int \dfrac{\sqrt{1 + (\tan \theta)^2}}{(\tan \theta)^4} (\sec \theta)^2\,d\theta = \int \dfrac{\sec \theta}{(\tan \theta)^4} (\sec \theta)^2\,d\theta = \int \dfrac{(\sec \theta)^3}{(\tan \theta)^4}\,d\theta =$$ \trigsubdisplay{$\left[x = \tan \theta, \quad dx = (\sec \theta)^2\,d\theta\right]$} {\epsfysize=1in \epsffile{intrev1.eps}}

$$\int \dfrac{1}{(\cos \theta)^3}\cdot \dfrac{(\cos \theta)^4}{(\sin \theta)^4}\,d\theta = \int \dfrac{\cos \theta}{(\sin \theta)^4}\,d\theta = \int \dfrac{\cos \theta}{u^4}\cdot \dfrac{du}{\cos \theta} = \int \dfrac{du}{u^4} =$$

$$\left[u = \sin \theta, \quad du = \cos \theta\,d\theta, \quad d\theta = \dfrac{du}{\cos \theta}\right]$$

$$-\dfrac{1}{3u^3} + C = -\dfrac{1}{3(\sin \theta)^3} + C = -\dfrac{1}{3}\cdot \dfrac{(1 + x^2)^{3/2}}{x^3} + C.\quad\halmos$$

Example. Compute $\displaystyle \int \ln (1 + x^2)\,dx$ .

$$\matrix{& \der {} x & & \displaystyle \int\,dx \cr & & & \cr + & \ln (1 + x^2) & & 1 \cr & & \searrow & \cr - & \dfrac{2x}{x^2 + 1} & \to & x \cr}$$

$$\int \ln (1 + x^2)\,dx = x \ln (1 + x^2) - 2 \int \dfrac{x^2}{x^2 + 1}\,dx = x \ln (1 + x^2) - 2\int \left(1 - \dfrac{1}{x^2 + 1}\right)\,dx =$$

$$x \ln (1 + x^2) - 2x + 2\tan^{-1} x + C.\quad\halmos$$

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Last updated: December 2, 2005

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