Chapter 8 Integration Techniques and Improper Integrals
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1 Chapter 8 Integration Techniques and Improper Integrals 8.1 Basic Integration Rules 8.2 Integration by Parts 8.4 Trigonometric Substitutions 8.5 Partial Fractions 8.6 Numerical Integration 8.7 Integration by Tables and Other Integration Techniques 8.8 Improper Integrals
2 8.2 Integration by Parts Summary: Common Integrals Using Integration by Parts 1. For integrals of the form x n e ax dx, x n sin(ax) dx, x n cos(ax) dx let u = x n and let dv = e ax dx, sin ax dx, or cos ax dx. 2. For integrals of the form x n ln x dx, x n arcsin(ax) dx, x n arctan(ax) dx let u = ln x, arcsin(ax), or arctan(ax) and let dv = x n dx. 3. For integrals of the form e ax sin(bx) dx or e ax cos(bx) dx let u = sin bx or cos(bx) and let dv = e ax dx.
3 In this section, we will study techniques for evaluating integrals of the form sin m x cos n x dx and sec m x tan n x dx Some Trigonometric Formulas Pythagorean identities sin 2 x + cos 2 x = 1, 1 + tan 2 x = sec 2 x, 1 + cot 2 x = csc 2 x Power-reducing Formulas sin 2 1 cos 2x x =, cos cos 2x x = 2 2 Product-to-Sum Formulas sin mx sin nx = 1 2 (cos sin mx cos nx = 1 2 (sin cos mx cos nx = 1 2 (cos m n x cos[ m + n x]) m n x + sin[ m + n x]) m n x + cos[ m + n x])
4 Basic Derivative and Antiderivative Formulas with Trig Functions d d sin u = u cos u, dx d cos u = u sin u, dx dx tan u = u sec2 u sin x dx = cos x + C, cos x dx = sin x + C, sec 2 x dx = tan x + C tan x dx = ln cos x + C, sec x dx = ln sec x + tan x + C sin(kx) dx = 1 cos(kx) + C, k cos(kx) dx = 1 sin(kx) + C, k sec 2 (kx) dx = 1 tan(kx) + C k
5 Examples Find sin 2 x dx, cos 2 x dx, tan 2 dx sin 3 x dx, cos 3 x dx, tan 3 dx
6 Guidelines for Evaluating Integrals Involving Powers of Sine and Cosine When the power of the sine is odd and positive, save one sine factor and convert the remaining factors to cosines. Then expand and integrate using the substitution u = cos x. sin 2k+1 x cos n x dx = sin 2 x k cos n x sin x dx = 1 cos 2 x k cos n x sin x dx When the power of the cosine is odd and positive, save one cosine factor and convert the remaining factors to sines. Then expand and integrate using the substitution u = sin x. sin m x cos 2k+1 x dx = sin m x cos 2 x k cos x dx = sin m x 1 sin 2 x k cos x dx When the powers of both the sine and cosine are even and nonnegative, make repeated use of the power-reducing formulas to convert the integrand to odd powers of the cosine. Then proceed as in the second guideline.
7 Example 1. Find sin 3 x cos 4 x dx, sin 4 x cos 3 x dx,
8 Example 2. Find cos 3 x sin x dx
9 Example 3. Find 0 π/2 cos 4 x dx Wallis Formulas. 1. If n is odd (n 3), then π/2cos n x dx = n If n is even (n 2), then π/2cos n x dx = n 1 0 n n π 2
10 Guidelines for Evaluating Integrals Using Product-to-Sum Formulas When the sign and cosine functions have different inputs with constant x, product-tosum formulas can be used sin mx sin nx = 1 2 (cos sin mx cos nx = 1 2 (sin cos mx cos nx = 1 2 (cos m n x cos[ m + n x]) m n x + sin[ m + n x]) m n x + cos[ m + n x]) Example 8. Find sin(5x) cos(4x) dx
11 Guidelines for Evaluating Integrals Involving Powers of Secant and Tangent When the power of the secant is even and positive, save a secant-squared factor and convert the remaining factors to tangents. Then expand and integrate using u = tan x. sec 2k x tan n x dx = sec 2 x k 1 tan n x sec 2 x dx = 1 + tan 2 x k 1 tan n x sec 2 x dx When the power of the tangent is odd and positive, save a secant-tangent factor and convert the remaining factors to secant. Then expand and integrate using u = sec x. sec m x tan 2k+1 x dx = sec m 1 x tan 2 x k sec x tan x dx = sec m 1 x sec 2 x 1 k sec x tan x dx When there are no secant factors and the power of the tangent is even and positive, convert a tangentsquared factor to a secant-squared factor, then expand and repeat if necessary tan n x dx = (tan n 2 x)(tan 2 x) dx = (tan n 2 x)(sec 2 x 1) dx When the integral is of the form sec m x dx where m is odd and positive, use integration by parts. When the first four guidelines do not apply, try converting to sines and cosines.
12 Example 4. Find tan 3 x sec x dx
13 Example 5. Find sec 4 (3x) tan 3 (3x) dx
14 Example 6. Evaluate 0 π/4 tan 4 x dx
15 Example 7. Evaluate sec x tan 2 x dx
16 8.4 Trigonometric Substitution In this section, we will study techniques for evaluating integrals involving the radicals a 2 u 2, a 2 + u 2, and u 2 a 2 using trigonometric substitution. Make sure we know the Pythagorean identities of trigonometric functions sin 2 x + cos 2 x = 1, 1 + tan 2 x = sec 2 x, 1 + cot 2 x = csc 2 x Note that the inverse hyperbolic functions can be used to find the integrals of the forms 1 u 2 ± a 2 du = ln u + u2 ± a 2 + C 1 u a 2 ± u du = 1 2 a ln a + a2 ± u2 + C u
17 8.4 Trigonometric Substitution Trigonometric Substitution a > 0 For integrals involving a 2 u 2, let u = a sin θ. Then a 2 u 2 = a 2 a 2 sin 2 θ = a 2 (1 sin 2 θ) = a 2 cos 2 θ = a cos θ where π 2 θ π 2 For integrals involving a 2 + u 2, let u = a tan θ. Then a 2 + ta u 2 = a 2 + a 2 tan 2 θ = a 2 (1 + tan 2 θ) = a 2 sec 2 θ = a sec θ where π 2 θ π 2 For integrals involving u 2 a 2, let u = a sec θ. Then u 2 a 2 = a 2 sec 2 θ a 2 = a 2 (sec 2 θ 1) = a 2 tan 2 θ a tan θ for u > a, where 0 θ < π/2 = a tan θ for u < a, where π/2 < θ π
18 8.4 Trigonometric Substitution Example 1 Find x x dx 2
19 8.4 Trigonometric Substitution Example 2 Find 1 4x dx
20 8.4 Trigonometric Substitution Example 3 Find 1 x /2 dx
21 8.4 Trigonometric Substitution Example 4 Evaluate 2 x x dx
22 8.4 Trigonometric Substitution Special Integration Formulas a > 0 a 2 u 2 du = 1 2 u a2 u 2 + a 2 arcsin u a + C u 2 a 2 du = 1 2 u a2 u 2 a 2 ln u + u 2 a 2 + C, u > a u 2 + a 2 du = 1 2 u u2 + a 2 + a 2 ln u + u 2 + a 2 + C
23 8.4 Trigonometric Substitution Example 5 Find the arc length of the graph of f x = 1 2 x2 from x = 0 to x = 1.
Example. Evaluate. 3x 2 4 x dx.
3x 2 4 x 3 + 4 dx. Solution: We need a new technique to integrate this function. Notice that if we let u x 3 + 4, and we compute the differential du of u, we get: du 3x 2 dx Going back to our integral,
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