- •Functions
- •The Concept of a Function
- •Trigonometric Functions
- •Inverse Trigonometric Functions
- •Logarithmic, Exponential and Hyperbolic Functions
- •Limits and Continuity
- •Introductory Examples
- •Continuity Examples
- •Linear Function Approximations
- •Limits and Sequences
- •Properties of Continuous Functions
- •The Derivative
- •The Chain Rule
- •Higher Order Derivatives
- •Mathematical Applications
- •Antidifferentiation
- •Linear Second Order Homogeneous Differential Equations
- •Linear Non-Homogeneous Second Order Differential Equations
- •Area Approximation
- •Integration by Substitution
- •Integration by Parts
- •Logarithmic, Exponential and Hyperbolic Functions
- •The Riemann Integral
- •Volumes of Revolution
- •Arc Length and Surface Area
- •Techniques of Integration
- •Integration by formulae
- •Integration by Substitution
- •Integration by Parts
- •Trigonometric Integrals
- •Trigonometric Substitutions
- •Integration by Partial Fractions
- •Fractional Power Substitutions
- •Numerical Integration
- •Integrals over Unbounded Intervals
- •Discontinuities at End Points
- •Improper Integrals
- •Sequences
- •Monotone Sequences
- •Infinite Series
- •Series with Positive Terms
- •Alternating Series
- •Power Series
- •Taylor Polynomials and Series
- •Applications
- •Parabola
- •Ellipse
- •Hyperbola
- •Polar Coordinates
- •Graphs in Polar Coordinates
- •Areas in Polar Coordinates
- •Parametric Equations
6.2. INTEGRATION BY SUBSTITUTION |
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csc t(csc t − cot t)dt |
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sin x |
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21. |
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dx |
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cos2 x |
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sin2 x |
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sin2 t |
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cos2 t |
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cos x |
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sin3 t − 3 |
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cos3 t + 2 |
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25. |
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tan2 t dt |
26. |
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cot2 t dt |
27. |
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28. |
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29. |
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sinh t dt |
30. |
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cosh t dt |
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31.Determine f(x) if f0(x) = cos x and f(0) = 2.
32.Determine f(x) if f00(x) = sin x and f(0) = 1, f0(0) = 2.
33.Determine f(x) if f00(x) = sinh x and f(0) = 2, f0(0) = −3.
34.Prove each of the integration formulas 1–77.
6.2Integration by Substitution
Theorem 6.2.1 Let f(x), g(x), f(g(x)) and g0(x) be continuous on an interval [a, b]. Suppose that F 0(u) = f(u) where u = g(x). Then
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f(g(x))g0(x)dx = Z |
f(u)du = F (g(x)) + C |
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u=g(b) |
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Zab f(g(x))g0(x)dx = Zu=g(u) |
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Proof. See the proof of Theorem 5.3.1.
Exercises 6.2 In problems 1–39, evaluate the integral by making the given substitution.
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CHAPTER 6. |
TECHNIQUES OF INTEGRATION |
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3x(x2 + 1)10dx, u = x2 + 1 |
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x sin(1 + x2)dx, u = 1 + x2 |
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dt, x = √ |
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3. |
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√t |
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(1 + x3)3/2 dx, u = 1 + x3 |
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cos( t) |
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3x |
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2earcsin x |
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3earccos x |
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dx, u = arcsin x |
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1 − x2 |
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x 4x2 dx, u = 4x2 |
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10sin x cos x dx, u = sin x |
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4arctan x |
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dx, u = 4arctan x |
10. |
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dx, u = 1 + ln x |
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5arcsec x |
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Z (tan 2x)3 sec2 2x dx, u = tan 2x |
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11. |
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x√ |
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dx, u = arcsec x |
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x2 − 1 |
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(cot 3x)5 csc2 3x dx, u = cot 3x |
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sin21 x cos x dx, u = sin x |
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15. |
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cos5 x sin x dx, u = cos x |
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(1 + sin x)10 cos x dx, u = 1 + sin x |
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sin3 x dx, u = cos x |
18. |
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cos3 x dx, u = sin x |
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tan3 x dx, u = tan x |
20. |
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cot3 x dx, u = cot x |
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sec4 x dx, u = tan x |
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csc4 x dx, u = cot x |
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sin3 x cos3 x dx, u = sin x |
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sin3 x cos3 x dx, u = cos x |
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25. |
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tan4 x dx, u = tan x |
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sin(ln x) |
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6.2. |
INTEGRATION BY SUBSTITUTION |
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27. |
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tan3 x sec4 x dx, u = sec x |
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x cos(ln(1 + x2)) |
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cot3 x csc4 x dx, u = csc x |
30. |
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√9 − x2 |
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√4 + x2 |
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4 + x2 , x = 2 tan t |
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4 − x2 , x = 2 tanh t |
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x√x2 − 4 |
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4esin(3x) cos(3x)dx, u = sin 3x |
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x 3(x2+4)dx, u = 3x2+4 |
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3 etan 2x sec2 x dx, u = tan 2x |
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39. |
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x + 2 |
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Evaluate the following definite integrals. |
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41. |
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(x + 1)30dx |
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2 x(4 − x2)1/2dx |
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43. |
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π/4 tan3 x sec2 x dx |
44. |
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1 x3(x2 + 1)3dx |
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(x + 1)(x − 2)10dx |
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8 x2(1 + x)1/2dx |
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π/6 sin(3x)dx |
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π/4 cos(2x)dx |
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49. |
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π/4 sin3 2x cos 2x dx |
50. |
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π/6 cos4 3x sin 3x dx |
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earctan x |
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1/2 |
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earcsin x |
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1 + x2 |
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1 − x2 |
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CHAPTER 6. TECHNIQUES OF INTEGRATION |
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earcsec x |
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53. |
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x√ |
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54. |
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x2 − 1 |
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6.3Integration by Parts
Theorem 6.3.1 Let f(x), g(x), f0(x) and g0(x) be continuous on an interval [a, b]. Then
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f(x)g0 |
(x)dx = f(x)g(x) − Z |
g(x)f0(x)dx |
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(x)dx = (f(b)g(b) − f(a)g(a)) − Zab g(x)f0(x)dx |
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udv = uv − Z |
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where u = f(x) and dv = g0(x)dx are the parts of the integrand. Proof. See the proof of Theorem 5.4.1.
Exercises 6.3 Evaluate each of the following integrals.
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x sin x dx |
2. |
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x cos x dx |
3. |
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x ln x dx |
4. |
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x ex dx |
5. |
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x 4x dx |
6. |
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x2 ln x dx |
7. |
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x2 sin x dx |
8. |
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x2 cos x dx |
9. |
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x2ex dx |
10. |
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x2 10x dx |
6.3. |
INTEGRATION BY PARTS |
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ex sin x dx (Let u = ex twice and solve.) |
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ex cos x dx (Let u = ex twice and solve.) |
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e2x sin 3x dx (Let u = e2x twice and solve.) |
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x sin(3x)dx |
15. |
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x2 cos(2x)dx |
16. |
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x2e4xdx |
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x3 ln(2x)dx |
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x sec2 x dx |
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x csc2 x dx |
20. |
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x sinh(4x)dx |
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x2 cosh x dx |
22. |
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x cos(5x)dx |
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sin(ln x)dx |
24. |
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cos(ln x)dx |
25. |
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x arcsin x dx |
26. |
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x arccos x dx |
27. |
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x arctan x dx |
28. |
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x arcsec x dx |
29. |
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arcsin x dx |
30. |
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arccos x dx |
31. |
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arctan x dx |
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Verify the following integration formulas:
278 |
CHAPTER 6. TECHNIQUES OF INTEGRATION |
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sinn(ax)dx = |
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(ax) cos(ax) |
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cosn(ax)dx = |
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xnexdx = xnex − n Z |
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xn−1 cos x dx |
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xn cos x dx = xn sin x − n Z |
xn−1 sin x dx |
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eax sin(bx)dx = |
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eax cos(bx) dx = |
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a2 + b2 |
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secn x dx = |
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secn−2 x dx, n = 1, n > 0 |
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cscn x dx = |
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cscn−2 x dx, n = 1, n > 0 |
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sin4 x dx |
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cos5 x dx |
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x3exdx |
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x4 sin x dx |
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x3 cos x dx |
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e2x sin 3x dx |
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e3x cos 2x dx |
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x5 ln x dx |
6.3. |
INTEGRATION BY PARTS |
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51. |
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sec3 x dx |
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csc3 x dx |
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tann x dx = |
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tann−1 x − Z |
tann−2 x dx, n 6= 1 |
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cotn x dx = |
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cotn−1 x − Z |
cotn−2 x dx, n 6= 1 |
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sin2n+1 x dx = − Z |
(1 − u2)ndu, u = cos x |
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56.cos2n+1 x dx = − (1 − u2)ndu, u = sin x
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sin2n+1 x cosm x dx = − Z |
(1 − u2)numdu, u = cos x |
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cos2n+1 x sinm x dx = Z (1 − u2)numdu, u = sin x |
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tann x sec2m x dx = |
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un(1 + u2)m−1du, u = tan x |
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cotn x csc2m x dx = − Z |
un(1 + u2)m−1du, u = cot x |
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tan2n+1 x secm x dx = Z (u2 − 1)num−1du, u = sec x |
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cot2n+1 x cscm x dx = − Z (u2 − 1)num−1du, u = csc x |
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64. |
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sin mx cos nx dx = −2 |
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