Math 6 Summer 05 Homework 5 Solutions Drew Armstrong Book Problems: Chap.5 Eercises, 8, Chap 5. Eercises 6, 0, 56 Chap 5. Eercises,, 6 Chap 5.6 Eercises 8, 6 Chap 6. Eercises,, 30 Additional Problems: A. Let r > 0 be constant. In this problem you will evaluate the following integral in two different ways: r r d (a) Interpret this integral as the area of a shape you know. (b) Use the substitution r sin θ and the trigonometric identities Solutions: r sin θ cos θ and cos θ cos(θ) +. Then use the substitution u θ..5.. Evaluate the integral 3 ( + ) 5 d using the substitution u +. Since u + we have du 3 d and hence 3 ( + ) 5 d 3 (u) 5 d u 5 3 d ( ) du u 5 u 5 du 6 u6 + C ( + ) 6 + C,.5.8. Evaluate the integral cos( 3 ) d. We will use the substitution u 3, so that du 3 d. Then we have cos( 3 ) d cos(u) d
cos(u) d ( ) du cos(u) 3 cos(u) du 3 3 sin(u) + C 3 sin(3 ) + C,.5.. Evaluate the integral ( +) d. We will use the substitution u +, so that du d. Then we have ( + ) d u d u d ( ) du u u du u + C u + C ( + ) + C, 5..6. Differentiate f() ln(). We use the product rule to compute f () ( ln() ) ( ln()) () ln() + (ln()) ln() + ln() + ln(). [Remark: Hey, we just discovered by accident that ln() d ln() + C.
That was lucky!] 5..0. Differentiate y ln(). First we write y (ln()). Then we use the chain rule to get dy d ( )(ln()) (ln()) (ln()) (ln()). 5..56. Evaluate the integral 3 0 d 5+. We use the substitution u 5 +, so that du 5 d. Then we have 3 0 d 5 + d 5... Evaluate the integral ( 5 + 5 ) d. 5 3 0 3 d u du/5 u 0 u6 u u du 5 ln u u6 u (ln(6) ln()) 5 0.555 Here we just have to remember or look up the rules: ( 5 + 5 ) d 5 d + 5... Evaluate the integral d. 5 d 6 6 + ln(5) 5 + C, Here we use the substitution u, so that du d. Then we have d u d u d ( ) du u u d ln() u + C
ln() + C, 5..6. Evaluate the integral + d. Here we use the substitution u +, so that du ln() d. Then we have + d u d u d ( ) du u ln() ln() u du ln u + C ln() ln + + C ln() ln( + ) + C ln() log ( + ) + C, necessary.] 5.6.8. Simplify the epression tan(sin ). There are two ways to do this problem. [Remark: The last two steps of simplification were not () Well, one thing we do know is that sin(sin ). [This is the definition of sin.] So we have tan(sin ) sin(sin ) cos(sin () cos(sin ). Now we have to compute cos(sin ). First we recall that cos θ + sin θ for any θ. Then we substitute θ sin to get Finally we have cos (sin ) + sin (sin ) tan(sin ) cos (sin ) + cos (sin ) cos(sin ). cos(sin ).
() Let θ sin, so that sin θ. Now let s draw a right angled tringle with angle θ and hypotenuse of length. Since sin θ, the length of the opposite side must be. Let? be the length of the adjacent side. The Pythagorean Theorem tells us that Finally, we have? +??. tan(sin ) tan θ opposite adjacent 5.6.6. Find the derivative of the function tan ( ).. First we have to remember the formula d d tan () +. [If we didn t remember the formula then we would have to rediscover it.] Then we use the chain rule to compute d d tan ( ) + ( ) d d +. 6... Evaluate the integral θ cos θ dθ using integration by parts, with u θ and dv cos θ dθ. Since u θ we have du dθ, and since dv cos θ dθ we have v sin θ. Then integration by parts gives udv uv vdu θ cos θ dθ θ sin θ sin θ dθ θ sin θ ( cos θ) + C θ sin θ + cos θ + C, 6... Evaluate the integral e θ cos(θ) dθ. We will use integration by parts with f(θ) cos(θ) and g (θ) e θ, so that f (θ) sin(θ) and g(θ) e θ. Then we have f(θ)g (θ) dθ f(θ)g(θ) f (θ)g(θ) dθ
e θ cos(θ) dθ e θ cos(θ) e θ sin(θ) dθ e θ cos(θ) e θ sin(θ) dθ. Did that help? Now we have to evaluate the integral e θ sin(θ) dθ. Okay, let s do it! Let F (θ) sin(θ) and G (θ) e θ, so that F (θ) cos(θ) and G(θ) e θ Then we have F (θ)g (θ) dθ F (θ)g(θ) F (θ)g(θ) dθ e θ sin(θ) dθ e θ sin(θ) ( e θ ) cos(θ) dθ e θ sin(θ) + e θ cos(θ) dθ Now we re back to where we started. But that s a good thing! Define A : e θ cos(θ) dθ. Putting our two equations together gives A e θ cos(θ) e θ sin(θ) dθ We conclude that A e θ cos(θ) ( e θ sin(θ) + A) A e θ cos(θ) + e θ sin(θ) A 5A e θ ( sin(θ) cos(θ)) A 5 e θ ( sin(θ) cos(θ)). e θ cos(θ) dθ 5 e θ ( sin(θ) cos(θ)) + C. [Remark: Good thing we didn t lose our confidence when the first integration by parts didn t work.] 6..30. First make a substitution and then use integration by parts to evaluate e d. First we let u, so that du d. Then we have e d u u e u du ue u du. Okay. Now we let f(u) u and g (u) e u, so that f (u) and g(u) e u. Then we have u u u f(u)g (u) f(u)g(u) u u f (u)g(u) du u ue u du ue u e u du (()e ()e ) (e e )
e e e + e e. And that s the answer. A. Compute the integral r r r d in two ways. (a) First we notice that this is just the area of a semicircle of radius r: Hence r r r d πr. (b) Second, we will follow the hints to evaluate the integral by hand. Let r sin θ, so that d r cos θ dθ. Then we have r d r r sin θ d r ( sin θ) d r cos θ d r cos θ d r cos θ(r cos θ dθ) r cos θ dθ ( r cos(θ) + ) dθ r (cos(θ) + ) dθ. Then we make the substitution u θ, so that du dθ, to get r (cos(θ) + ) dθ r (cos(u) + ) dθ r (cos(u) + ) du r (sin(u) + u) + C, Finally, since r sin θ we note that goes from r to r as θ goes from π/ to π/; and since u θ we note that θ goes from π/
to π/ as u goes from π to π. We conclude that r θπ/ r d r (cos(θ) + ) dθ r r r r r θ π/ uπ u π r [π] πr. Which method do you prefer? (sin(u) + u) (cos(u) + ) du uπ u π [(sin(π) + π) (sin( π) + ( π))] [(0 + π) (0 π)] [Remark: That was the final homework problem of the course. Now we have come full circle. On HW Problem we discussed Archimedes proof that the area of a circle is πr. Here we used the methods of Calculus to come up with a completely different proof. Calculus can be used to solve a wide array of problems. And once you have some practice, it doesn t really require that much effort. We can all be Archimedes now.]