TRIGONOMETRIC IDENTITIES

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1 Integration TRIGONOMETRIC IDENTITIES Graham S McDonald and Silvia C Dalla A self-contained Tutorial Module for practising integration of expressions involving products of trigonometric functions such as sin nx sin mx Table of contents Begin Tutorial c 004 g.s.mcdonald@salford.ac.uk

2 . Theory. Exercises 3. Answers 4. Standard integrals 5. Tips Full worked solutions Table of contents

3 Section : Theory 3. Theory Integrals of the form sin nx sin mx, and similar ones with products like sin nx cos mx and cos nx cos mx, can be solved by making use of the following trigonometric identities: sin A sin B = [cos (A + B) cos (A B)] sin A cos B = [sin (A + B) + sin (A B)] cos A cos B = [cos (A + B) + cos (A B)] Using these identities, such products are expressed as a sum of trigonometric functions This sum is generally more straightforward to integrate

4 Section : Exercises 4. Exercises Click on EXERCISE links for full worked solutions (9 exercises in total). Perform the following integrations: Exercise. cos 3x cos x dx Exercise. sin 5x cos 3x dx Exercise 3. sin 6x sin 4x dx Theory Standard integrals Answers Tips

5 Section : Exercises 5 Exercise 4. cos ωt sin ωt dt, where ω is a constant Exercise 5. cos 4ωt cos ωt dt, where ω is a constant Exercise 6. sin x dx Exercise 7. sin ωt dt, where ω is a constant Exercise 8. cos t dt Theory Standard integrals Answers Tips

6 Section : Exercises 6 Exercise 9. cos kx dx, where k is a constant Theory Standard integrals Answers Tips

7 Section 3: Answers 7 3. Answers. 0 sin 5x + sin x + C,. 6 cos 8x 4 cos x + C, 3. 0 sin 0x + 4 sin x + C, 4. 6ω cos 3ωt + ω cos ωt + C, 5. ω sin 6ωt + 4ω sin ωt + C, 6. 4 sin x + x + C, 7. 4ω sin ωt + t + C, 8. 4 sin t + t + C, 9. 4k sin kx + x + C.

8 Section 4: Standard integrals 8 4. Standard integrals f (x) x n x f(x)dx f (x) f(x)dx xn+ n+ (n ) [g (x)] n g (x) ln x g (x) g(x) [g(x)] n+ n+ (n ) ln g (x) e x e x a x ax ln a (a > 0) sin x cos x sinh x cosh x cos x sin x cosh x sinh x tan x ln cos x tanh x ln cosh x cosec x ln tan x cosech x ln tanh x sec x ln sec x + tan x sech x tan e x sec x tan x sech x tanh x cot x ln sin x coth x ln sinh x sin x cos x x x sin x 4 sinh x sinh x 4 x + sin x 4 cosh x sinh x 4 + x

9 Section 4: Standard integrals 9 f (x) f (x) dx f (x) f (x) dx a +x a tan x a a x a ln a+x a x (0< x <a) (a > 0) x a a ln x a x+a ( x > a>0) a x sin x a a +x ( a < x < a) x a ln ln x+ a +x a x+ x a a (a > 0) (x>a>0) a x a [ sin ( ) x a a +x a ] x + x a x a a a [ [ sinh ( x a cosh ( x a ) + x a +x a ] ) + x ] x a a

10 Section 5: Tips 0 5. Tips STANDARD INTEGRALS are provided. Do not forget to use these tables when you need to When looking at the THEORY, STANDARD INTEGRALS, AN- SWERS or TIPS pages, use the Back button (at the bottom of the page) to return to the exercises Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct Try to make less use of the full solutions as you work your way through the Tutorial

11 Solutions to exercises Full worked solutions Exercise. cos 3x cos x dx: Use cos A cos B = [cos (A + B) + cos (A B)] i.e. taking A = 3x and B = x: cos 3x cos x dx = [cos (3x + x) + cos (3x x)] [cos 5x + cos x] dx = Each term under the integration sign is a function of a linear function of x, i.e. f(ax+b) dx = a f(u)du, where u = ax+b, du = a dx, i.e. dx = du a. i.e. cos 3x cos x dx = 5 sin 5x+ sin x+c = 0 sin 5x+ sin x+c. Return to Exercise

12 Solutions to exercises Exercise. sin 5x cos 3x dx: Use sin A cos B = [sin (A + B) + sin (A B)] i.e. taking A = 5x and B = 3x: sin 5x cos 3x dx = [sin (5x + 3x) + sin (5x 3x)] = [sin 8x + sin x] dx i.e. sin 5x cos 3x dx = 8 cos 8x cos x + C = 6 cos 8x cos x + C. 4 Return to Exercise

13 Solutions to exercises 3 Exercise 3. sin 6x sin 4x dx: Use sin A sin B = [cos (A + B) cos (A B)] i.e. taking A = 6x and B = 4x: sin 6x sin 4x dx = = [cos (6x + 4x) cos (6x 4x)] [cos 0x cos x] dx i.e. sin 6x sin 4x dx = 0 sin 0x + sin x + C = 0 sin 0x + sin x + C. 4 Return to Exercise 3

14 Solutions to exercises 4 Exercise 4. cos ωt sin ωt dt: Use sin A cos B = [sin (A + B) + sin (A B)] i.e. taking A = ωt and B = ωt: cos ωt sin ωt dt = [sin (ωt + ωt) + sin (ωt ωt)] = [sin 3ωt + sin ( ωt)] dt = [sin 3ωt sin ωt] dt = 3ω cos 3ωt + cos ωt + C ω = cos 3ωt + cos ωt + C. 6ω ω Return to Exercise 4

15 Solutions to exercises 5 Exercise 5. cos 4ωt cos ωt dt: Use cos A cos B = [cos (A + B) + cos (A B)] i.e. taking A = 4ωt and B = ωt: cos 4ωt cos ωt dt = = [cos (4ωt + ωt) + cos (4ωt ωt)] [cos 6ωt + cos ωt] dt = 6ω sin 6ωt + sin ωt + C ω = sin 6ωt + sin ωt + C. ω 4ω Return to Exercise 5

16 Solutions to exercises 6 Exercise 6. sin x dx: For the particular case: A=B=x, the formula: sin A sin B = [cos (A + B) cos (A B)], reduces to: sin x = (cos x ) (a half-angle formula) i.e. sin x dx = (cos x ) dx = sin x + x + C = 4 sin x + x + C. Return to Exercise 6

17 Solutions to exercises 7 Exercise 7. sin ωt dt: For the particular case: A=B=ωt, the formula: sin A sin B = [cos (A + B) cos (A B)], reduces to: sin ωt = (cos ωt ) i.e. sin ωt dt = (cos ωt ) dt = ω sin ωt + t + C = 4ω sin ωt + t + C. Return to Exercise 7

18 Solutions to exercises 8 Exercise 8. cos t dt: For the particular case: A=B=t, the formula: cos A cos B = [cos (A + B) + cos (A B)], reduces to: cos t = (cos t + ) i.e. cos t dt = (cos t + ) dt = sin t + t + C = 4 sin t + t + C. Return to Exercise 8

19 Solutions to exercises 9 Exercise 9. cos kx dx: For the particular case: A=B=kx, the formula: cos A cos B = [cos (A + B) + cos (A B)], reduces to: cos kx = (cos kx + ) i.e. cos kx dx = (cos kx + )dx = k sin kx + x + C = 4k sin kx + x + C. Return to Exercise 9

Integration ALGEBRAIC FRACTIONS. Graham S McDonald and Silvia C Dalla

Integration ALGEBRAIC FRACTIONS Graham S McDonald and Silvia C Dalla A self-contained Tutorial Module for practising the integration of algebraic fractions Table of contents Begin Tutorial c 2004 g.s.mcdonald@salford.ac.uk