Trigonometric substitutions (8.3).

Review for Eam 2. 5 or 6 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Eam covers: 7.4, 7.6, 7.7, 8-IT, 8., 8.2. Solving differential equations (7.4). Inverse trigonometric functions (7.6). Hperbolic functions (7.7). Integration techniques (8-IT). Integration b parts (8.). Trigonometric integrals (8.2). Section not covered: Trigonometric substitutions (8.3). Review for Eam 2. 5 or 6 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Eam covers: 7.4, 7.6, 7.7, 8-IT, 8., 8.2. Solving differential equations (7.4). Inverse trigonometric functions (7.6). Hperbolic functions (7.7). Integration techniques (8-IT). Integration b parts (8.). Trigonometric integrals (8.2). Section not covered: Trigonometric substitutions (8.3).

Solving differential equations (7.4) Remark: Tpical problems in this section: () Find the function solution of = sin() 4 and (0) = 2. (2) The intensit L() of light feet beneath the surface of the ocean satisfies the equation L = kl, for some k > 0. If diving at 5 ft cuts the light intensit in half, how deep the light intensit falls below /8 the intensit at the surface? Solving differential equations (7.4) Eample Find the function solution of = sin() 4 Solution: 4 = sin() 4() () d = and (0) = 2. sin() d. The substitution u = (), with du = () d, implies 4 u du = sin() d 2u 2 = cos() + c, Therefore, 2 () = ( cos() + c)/2. The condition (0) < 0, implies () = c cos()/ 2. Furthermore, c 2 = 2 = c c = 5. 2 We conclude that = 5 cos()/ 2.

Solving differential equations (7.4) Eample The intensit L() of light feet beneath the surface of the ocean satisfies the equation L = kl, for some k > 0. If diving at 5 ft cuts the light intensit in half, how deep the light intensit falls below /8 the intensit at the surface? Solution: Integrate the differential equation, L () d = k d, u = L(), du = L () d L() du u = k d ln(u) = k + c L() = e k+c. Since L() = e k e c, calling L 0 = e c, we get the solution L() = L 0 e k. Solving differential equations (7.4) Eample The intensit L() of light feet beneath the surface of the ocean satisfies the equation L = kl, for some k > 0. If diving at 5 ft cuts the light intensit in half, how deep the light intensit falls below /8 the intensit at the surface? Solution: Recall: L() = L 0 e k. Now the first condition implies L 0 2 = L(5) = L 0 e 5k e 5k = 2 5k = ln(2) so we conclude that k = ln(2)/5. The second condition implies L 0 8 = L 0 e k e k = 8 k = ln(8). Using the value k = ln(2)/5, we get = ln(8) 5 ln(2) = 3(5) = 45.

Review for Eam 2. Eam covers: 7.4, 7.6, 7.7, 8-IT, 8., 8.2. Solving differential equations (7.4). Inverse trigonometric functions (7.6). Hperbolic functions (7.7). Integration techniques (8-IT). Integration b parts (8.). Trigonometric integrals (8.2). Section not covered: Trigonometric substitutions (8.3). Inverse trigonometric functions (7.6) Notation: In the literature is common the notation sin = arcsin, and similar for the rest of the trigonometric functions. Do not confuse sin() sin () = arcsin(). Remark: sin, cos have simple values at particular angles. θ sin(θ) cos(θ) 0 0 π/6 /2 3/2 π/4 2/2 2/2 π/3 3/2 /2 π/2 0

Inverse trigonometric functions (7.6) Remark: On certain domains the trigonometric functions are invertible. = sin() = cos() = tan() 0 π = csc() = sec() = cot() 0 0 π 0 π Inverse trigonometric functions (7.6) Remark: The graph of the inverse function is a reflection of the original function graph about the = ais. = arcsin() π = arccos() = arctan() 0 = arccsc() = arcsec() = arccot() π π 0 0 0

Inverse trigonometric functions (7.6) Theorem The derivative of inverse trigonometric functions are: arcsin () = 2, arccos () =,, 2 arctan () = + 2, arccot () = arcsec () = Recall arctan () = 2, + 2, R, arccsc () =, 2. tan ( arctan() ), tan () = cos2 () + sin 2 () cos 2 () tan () = + tan 2 (), = arctan(), arctan () = + 2. Inverse trigonometric functions (7.6) Remark: Tpical problems in this section: () Sketch the graphs of () = sec(), z() = arcsec(). State the respective domains and ranges. (2) Evaluate cos ( arcsin(/ (2) ). (3) Evaluate sec ( arctan( 2/3) ). (4) Find for () = arctan(3 2 ). d (5) Find I =. 2 2

Inverse trigonometric functions (7.6) Eample Evaluate sec ( arctan( 2/3) ). Solution: We onl need the relation between sec and tan, sec 2 (θ) = tan 2 (θ) +. Then holds sec(θ) = ± tan 2 (θ) +. We need to find the correct sign: θ = arctan( 2/3) (, 0). Since sec(θ) = / cos(θ), we conclude that sec(θ) > 0. Hence ( sec(arctan 2 ) ( ) = tan 3 2 (arctan 2 ) 4 3 ) + = 3 9 + = 9. We conclude that sec(arctan( 2/3)) = 3/3. Review for Eam 2. Eam covers: 7.4, 7.6, 7.7, 8-IT, 8., 8.2. Solving differential equations (7.4). Inverse trigonometric functions (7.6). Hperbolic functions (7.7). Integration techniques (8-IT). Integration b parts (8.). Trigonometric integrals (8.2). Section not covered: Trigonometric substitutions (8.3).

Hperbolic functions (7.7) Definition The complete set of hperbolic trigonometric functions is given b cosh() = e + e, sinh() = e e, 2 2 tanh() = sinh() cosh(), cosh() coth() = sinh(), csch() = sinh(), sech() = cosh(). Theorem The following identities hold, cosh 2 () sinh 2 () =, sinh(2) = 2 sinh() cosh(), cosh(2) = cosh 2 () + sinh 2 (), cosh 2 () = 2 [ + cosh(2) ], sinh 2 () = 2 [ + cosh(2) ]. Hperbolic functions (7.7) Remark: Tpical problems in this section: () Prove the identities: cosh 2 () sinh 2 () =, and cosh(2) = cosh 2 () + sinh 2 (), sinh(2) = 2 sinh() cosh()), cosh 2 () = 2 ( + cosh(2) ), sinh 2 () = 2 ( + cosh(2) ). (2) Know the derivatives and integrals of hperbolic functions.

Review for Eam 2. Eam covers: 7.4, 7.6, 7.7, 8-IT, 8., 8.2. Solving differential equations (7.4). Inverse trigonometric functions (7.6). Hperbolic functions (7.7). Integration techniques (8-IT). Integration b parts (8.). Trigonometric integrals (8.2). Section not covered: Trigonometric substitutions (8.3). Sections 8-IT, 8., 8.2 Remark: Evaluate the following integrals: ( + ) d (). (7) cot 3 () d. 2 2 8 d (2) 2 2 + 50. (8) sin 4 () d. (3) 3 (9) 3 cos() d. ln() d. π/2 (4) 2 e 2 d. (0) cos(2) d. d π/3 (5). sec 2 () () 8 2 π/4 tan() d. d (6), < 5. 2 ln() 25 2 (2) d.

Sections 8-IT, 8., 8.2 Remark: Evaluate the following integrals: () ( + ) d 2 2. Split the integral and do two substitutions. (2) (3) (4) (5) (6) 8 d 2 2 + 50. Complete the square and recall the arctan. 3 ln() d. 2 e 2 d. Three integrations b parts. Two integrations b parts. d 8 2. Complete the square and recall arcsin. d 25 2, < 5. Substitution and recall arcsin. Sections 8-IT, 8., 8.2 Remark: Evaluate the following integrals: (7) cot 3 () d. Write using sin, cos and substitution. (8) sin 4 () d. Double angle formula, twice. (9) 3 cos() d. Integrations b parts, three times. (0) () (2) π/2 π/3 π/4 2 ln() cos(2) d. sec 2 () tan() d. d. Double angle formula, cancel Write using sin and cos, and substitution. Substitution..

Sections 8-IT, 8., 8.2 Eample Evaluate I = ( + ) d 2 2. Solution: Split the integral: I = d + 2 2 d 2 2. For the first integral substitute = 2, then d = 2 d. d I = = d = arcsin( 2 ) + c. 2 2 2 2 2 For the second integral substitute u = 2 2, then du = 4 d. I 2 = du = 4 u 4 (2 u) + c = 2 2 2 + c. We conclude: I = 2 arcsin( 2 ) 2 2 2 + c. Sections 8-IT, 8., 8.2 Eample Evaluate I = d 8 2. Solution: Complete the square and recall arcsin. d I = 2 + 2(4) = d 2 + 2(4) 4 2 + 4, 2 I = d 4 2 ( 2 2(4) + 4 2 ) = d 4 2 ( 4) 2 I = 4 d [( 4)/4] 2. Substitute u = ( 4)/4, then du = d/4. du ( ( 4) ) I = I = arcsin(u) + c I = arcsin + c. u 2 4

Sections 8-IT, 8., 8.2 Eample Evaluate I = π/2 cos(2) d. Solution: Double angle formula, cancel Recall: sin 2 (θ) = [ cos(2θ)]/2. Hence, I = π/2 2 sin 2 () d = 2 Since sin() < 0 for (, 0), I = 2 0 I = 2 cos() 0 2 cos() sin() d + 2 π/2 0. π/2 π/2 0 sin() d. sin() d. = 2( 0) 2(0 ) = 2 2.