Differential and Integral Calculus

Transcription

1 School of science an engineering El Akhawayn University Monay, March 31 st, 2008

2 Outline 1

3 Definition of hyperbolic functions: The hyperbolic cosine an the hyperbolic sine of the real number x are enote by cosh x an sinh x an are efine to be cosh x = ex + e x 2 an sinh x = ex e x. 2 The other hyperbolic functions are efine by analogy with trigonometry, as follows The hyperbolic tangent tanh x = sinh x cosh x = ex e x e x + e x, The hyperbolic cotangent coth x = cosh x sinh x = ex + e x e x, x 0. e x

4 Definition of hyperbolic functions The hyperbolic secant sech x = 1 cosh x = 2 e x + e x, The hyperbolic cosecant csch x = 1 sinh x = 2 e x, x 0. e x Ientities: The hyperbolic functions have "similar" ientities as the trigonometric functions Trigonometric Functions Ientities Ientities cos 2 x + sin 2 x = 1 cosh 2 x sinh 2 x = tan 2 x = sec 2 x 1 tanh 2 x = sech 2 x cot 2 x + 1 = csc 2 x coth 2 x 1 = csch 2 x

5 Ientities: (Continue) Other hyperbolic ientities are given by sinh(x + y) = sinh x cosh y + cosh x sinh y cosh(x + y) = cosh x cosh y + sinh x sinh y sinh 2x = 2 sinh x cosh x cosh 2x = cosh 2 x + sinh 2 x cosh 2 x = 1 2 (cosh 2x + 1) sinh 2 x = 1 2 (cosh 2x 1) Example 1: Using the efinitions of cosh x an sinh x, we get cosh 2 x sinh 2 x = 1 4 (ex + e x ) (ex e x ) 2 = 1 4 (e2x e 2x ) 1 4 (e2x 2 + e 2x ) = 1.

6 Derivatives an Integrals of : By efinition of the cosine hyperbolic function, we have x cosh x = ( 1 x 2 ex + 1 ) 2 e x = 1 2 ex 1 2 e x = sinh x If u is a ifferentiable function of x, then The other formulas are given by sinh u = (cosh u)u x x x coth u = ( csch2 u) u x csch u = ( csch u coth u)u x x cosh u = (sinh u)u x x. x tanh u = (sech2 u) u x sech u = ( sech u tanh u)u x x

7 Example 2: Differentiate the following functions f(x) = sinh 2 3x; g(x) = x tanh x; h(x) = sech (x 2 ). Solution: Using the Chain rule, we obtain D x f(x) = 6 sinh 3x cosh 3x D x g(x) = tanh x + xsech 2 x D x h(x) = 2xsech(x 2 ) tanh(x 2 ) Antierivatives of hyperbolic functions: The antierivatives of hyperbolic functions, are given by sinh uu = cosh u + C; cosh uu = sinh u + C.

8 Antierivatives of hyperbolic functions: sech 2 uu = tanh u + C, sech u tanh uu = sech u + C, csch 2 uu = coth u + C csch u coth uu = csch u + C. Example 3: Evaluate the following integrals sinh 4xx, sinh x cosh xx. Solution: Let u = 4x, then u = 4x, so sinh 4xx = 1 sinh uu = cosh u + C = 1 cosh 4x + C. 4

9 Solution: (Continue) By substitution with u = sinh x, then u = cosh xx, hence sinh x cosh xx = uu = 1 2 u2 + C = 1 2 sinh2 x + C. Example 4: Evaluate the integrals cosh 2 xx; 1 0 tanh 2 xx. Solution: We have cosh 2 xx = 1 2 (cosh 2x + 1) = 1 4 sinh 2x + x 2 + C.

10 Solution: (Continue) Using hyperbolic functions ientities, we get 1 0 tanh 2 xx = 1 0 (1 sech 2 x)x = [x tanh x] 1 0 = 1 tanh 1 = 1 e e 1 e + e 1 = 2e 1 e + e 1 = 2 e Example 5: Evaluate the integral coth xcsch 2 xx.

11 Solution: Let u = coth x. Then u = csch 2 xx. So coth xcsch 2 xx = uu = 1 2 u2 + C = 1 2 coth2 x + C. Example 6: Evaluate the inefinite integral 1 (e x + e x ) 2 x.

12 Solution: We have 1 (e x + e x ) 2 x = 1 ( 2 4 e x + e x = 1 4 ) 2 x sech 2 xx = 1 tanh x + C. 4