Math 102 Spring 2008: Solutions: HW #3 Instructor: Fei Xu

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1 Math Spring 8: Solutions: HW #3 Instructor: Fei Xu. section 7., #8 Evaluate + 3 d. + We ll solve using partial fractions. If we assume 3 A + B + C, clearing denominators gives us A A + B B + C +. Then associating like terms gives us that A + C, A + B, and B. Therefore, B, A, and C. Then we have + 3 d (. section 7., #3 Evaluate ) d ln ln + + K. + ( + ) ( + ) d. Again, we ll use partial fractions. If we assume that + A+B + + C+D + ( +) ( +) + E+F ( +), then clearing denominators gives us that A( ) + B( + + ) + C( ) + D( ) + E( 3 + ) + F ( + ) (A + C) + (B + D) + (A + 3C + E) 3 (B + 3D + F ) + + (A + C + E) + (B + D + F ) Therefore, we have that A + C, B + D, A + 3C + E B + 3D + F, A + C + E, B + D + F Solving for this system, we get that A, B, C, D, E, and F 3. Therefore, ( + ( + ) ( + ) d + ) ( + ) d arctan ( ) arctan + 3 ( arctan + ) + C +

2 For the last integral on the right, we use trigonometric substitution, let u tan θ, then 3 3 sec ( + ) d θ sec θ dθ 3 sec θ dθ 3 (θ + sin θ cos θ) 3 ( arctan + ) + 3. section 7., #38 Make a preliminary substitution before using the method of partial fractions to evaluate cos θ sin θ sin θ 6 d Let s use the substitution u sin θ, so du cos θ dθ. Then we have cos θ sin θ sin θ 6 d u u 6 du Then using partial fractions, let u u 6 A (A + B)u + (A 3B), and so A, B cos θ sin θ sin θ 6 d u 3 + u u 6 du ( u 3 u + B u+, so we get that. Then we have ) du (ln u 3 ln u + ) + C (ln sin θ 3 ln sin θ + + C. section 7.6, # Use trigonometric substitution to evaluate d 9 Let 3 sin u. Then we have 9 sin u ( sin u) cos u

3 and d 3 cos u du. This gives us that d u + C cos u du cos u 3 arcsin 3 + C. section 7.6, # Use trigonometric substitution to evaluate 3 + d Let tan u, so we have that + tan u + ( sec u ) + sec u and du sec u du. This gives us that 3 tan 3 + d u sec u du sec u tan 3 u sec u du (sec u ) tan u sec u du Now make the substitution w sec u, so dw sec u tan u du. Then we have 3 (sec + d u ) tan u sec u du (w ) dw ( ) 3 w3 w + C ( ) 3 sec3 u sec u + C Note that the right triangle with angle u, opposite side, and adjacent side has hypotenuse +. Therefore, we have that sec u +, so we get ( ( 3 + d 3 + ) 3/ ( + ) ) / + C ( + ) ( / ( + ) ) 3 ( + ) / ( ) + C 3

4 6. section 7.6, #6 Use trigonometric substitution to evaluate + d Let tan u, so we have d sec u du and + + tan u + (sec u ) sec u Then we have + d sec 3 u du sec u tan u + ln sec u + tan u + C from integral formula for secant. Now note that the right triangle with angle u, opposite side, and adjacent side has hypotenuse +, so sec u + and tan u, and so + d sec u tan u + ln sec u + tan u + C + + ln C 7. section 7.6, # Use trigonometric substitution to evaluate ( ) d Let sin u, so sin u cos u and d cos u du. Then ( ) d cos u 6 cos u du sec 3 u du 8 6 sec u tan u + ln sec u + tan u + C 6 Now consider the right triangle with angle u, opposite side, and hypotenuse. This has adjacent side, so sec u and tan u. Then we get that ( ) d ln + + C 8( ) + 3 ln ( + ) ( + )( ) + C 8( ) + 3 ln + + C

5 8. section 7.6, #36 Use trigonometric substitution to evaluate ( ) 3/ d Let sec u, so d sec u tan u du, and sec u tan u. Therefore, ( ) ( ) 3/ d 3/ tan 3 u sec u tan u du tan u sec u du (sec u sec u + ) sec u du (sec u sec 3 u + sec u ) du From secant integral formula, we have sec u du sec3 u tan u + 3 sec 3 u du sec3 u tan u sec u tan u + 3 ln sec u + tan u + C 8 Then ( ) 3/ d (sec u sec 3 u + sec u ) du ( sec3 u tan u 8 sec u tan u + 3 ) ln sec u + tan u + C 8 Now consider the right triangle with angle u, adjacent side, and hypotenuse. It has opposite side, and so sec u and tan u. This gives us ( ) 3/ d ( ln + ) + C ln + + C 9. section 7.6, # Compute the arc length of the parabola y over the interval [, ].

6 With y, we have ds + (dy/d) d + d so we want to calculate L + d. Therefore, let tan θ, so then + + tan θ sec θ, and d sec θ. Now we have. section 7.7, #6 L + d sec θ sec θdθ sec 3 θdθ (sec θ tan θ + ln sec θ + tan θ ) [ ( + + ln + )] + [ ( + ln + )] The derivative of 3 is 3. So + 3 d d Now we complete the square to get 3 (+). Substituting u + we get d 3 du u Substituting v u/ we get dv v sin (v) + C du (u/) where to get this last equality we either used the substitution v sin θ or remembered the integral from memory. Substituting back we get so the final answer is sin (u/) + C sin ( + ) + C 3 + sin ( + ) + C.

7 . section 7.7, # One can do this integral by integration by parts. But let s do it by substituting sin θ. We get sin 3 θ (cos cos θdθ θ) Now sin 3 θ sin θ( cos θ) so we get sin θ cos 7 θ sin θ cos θ dθ sin 3 θ cos 7 θ dθ. Now we substitute u cos θ to get du u 7 du u 6u 6 u + C Substituting back we get 6 cos 6 θ cos θ + C 6(cos sin ) 6 (cos sin ) Now cos sin () so we get 6( ) 3 ( ).. section 7.7, #36 du a cos θdθ so the left side equals (a a sin a cos θdθ θ) n a n cos n a cos θdθ θ a n cos n θ dθ a n sec n θdθ 3. section 7.8, # The function must calculate so is undefined at in the interval [, ]. Hence we d and d. Now if > then > du/ d / ln(u) + C / ln( ) + C u where we substituted u. If < then < so d / ln( ) + C

8 . Now as b we have + and hence ln( ). Similarly, as a + we have + and hence ln( ). Hence d + So the integral does not converge.. section 7.8 #3 d lim b [/ ln( )] b + lim a [/ ln( )] a ( ) + (/ ln(3) ( ) cos as π/ so the problem is at π/. Now letting u cos we get sin du d (cos ) /3 u /3 3u /3 + C 3(cos ) /3 + C. Now as π/ we have cos() + and (cos ) /3. Thus the integral does not converge.. section 7.8 # + > so we only have to figure out what happens as ±. Now + + d tan () + ln( + ) + C which does not converge since ln( + ) as. On the other hand t t + + d [tan () + ln( + )] t t tan (t) tan ( t) + ln( + t ) ln( + ( t) ) tan (t) tan ( t) Now as t we have tan (t) π/ and tan ( t) π/ which means t + lim d π/ ( π/) π. t t +