Mathematics 114Q Integration Practice Problems SOLUTIONS. = 1 8 (x2 +5x) 8 + C. [u = x 2 +5x] = 1 11 (3 x)11 + C. [u =3 x] = 2 (7x + 9)3/2

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1 Mathematics 4Q Name: SOLUTIONS. (x + 5)(x +5x) 7 8 (x +5x) 8 + C [u x +5x]. (3 x) (3 x) + C [u 3 x] 3. 7x +9 (7x + 9)3/ [u 7x + 9] 4. x 3 ( + x 4 ) /3 3 8 ( + x4 ) /3 + C [u + x 4 ] 5. e 5x+ 5 e5x+ + C [u 5x + ]

2 Math 4Q 6. 4 cos(3x) 4 3 sin(3x)+c [u 3x] 7. sin(ln x) x cos(ln x)+c [u ln x] 8. x + x +4x 3 ln x +4x 3 + C [u x +4x 3] 9. (3 x + )( x ) ln 3 3x + + C [u x + ]. x ln x ln ln x + C [u ln x]. cos(5x) esin(5x) 5 e sin(5x) + C [u sin(5x)]

3 Math 4Q. π x sin(x ) Let u x. Then du x, so π x sin(x ) π x π x sin(x ) x sin(u)du ( cos(x ) ) π 3. x 4 x [Hint: If u 4 x, what does that make x in terms of u?] If u 4 x, then x 4 u and so du. all of this into the integral: Now just substitute x 4 x (4 u) u ( )du ( 4u / + u 3/) du 8 3 (4 x)3/ + 5 (4 x)5/ + C

4 Math 4Q 4. xe x xe x e x + C [u x, dv e x ] 5. x sin x x cos(x) + sin(x)+c [u x, dv sin(x)] 6. x ln x x ln x x 4 + C [u ln x, dv x] 7. ln x x ln x x + C [u ln x, dv ] 8. ln x x 5 4 ln x x 4 6 x 4 + C [u ln x, dv x 5 ]

5 Math 4Q 9. x e 3x You will have to use integration by parts twice. First use u x and dv e 3x. This will give you x e 3x x 3 e3x 3 xe 3x. For the new integral, use u x and dv e 3x. answer: This gives the x e 3x x 3 e3x [ x 3 3 e3x 3 ] e ex x 3 e3x x 9 e3x + 7 e3x + C. x 3 ln(5x) x4 4 ln(5x) x4 6 + C [u ln(5x), dv x 3 ]. x x +3 Let u x and dv (x + 3) /. Then x x +3 3 x(x + 3)3/ 3 (x + 3)3/ 3 x(x + 3)3/ 4 5 (x + 3)5/ + C.

6 Math 4Q. sin (x) [Hint: write sin (x) as sin(x) sin(x) and use the Pythagorean Theorem.] As instructed in the hint, write sin (x) sin(x) sin(x). and dv sin(x). Then Let u sin(x) sin (x) sin(x) cos(x)+ cos (x) sin(x) cos(x)+ ( sin (x) ) sin(x) cos(x)+x sin (x) Now add sin (x) to both sides and divide by to get sin (x) sin(x) cos(x)+x. 3. x sin(x) cos(x) [Hint: let u x. You will need to use the result of the previous problem.] Let u x and dv sin(x) cos(x). Then v sin (x), so x sin(x) cos(x) x sin (x) sin (x) Note that we used x sin (x) [ ] sin(x) cos(x)+x + C sin (x), which we obtained in the previous problem. 4. x cos(x) x sin(x) + cos(x)+c [u x, dv cos(x)]

7 Math 4Q 5. x cos(x) You will need to do integration by parts twice. The first time, let u x and dv cos(x). Then x cos(x) x sin(x) sin(x) x. Next, let u x and dv sin(x). Then x cos(x) x sin(x) x sin(x) x sin(x) ( x cos(x)+ cos(x) ) x sin(x)+x cos(x) sin(x)+c. 6. e x cos(x) [Hint: Do integration by parts twice. equation.] Look carefully at both sides of the resulting Let u e x and dv cos(x). Then e x cos(x) e x sin(x) e x sin(x). To solve this new integral, use u e x and dv sin(x). Then e x cos(x) e x sin(x) [ e x cos(x) e x( cos(x) ) ] Now add e x cos(x) e x sin(x)+e x cos(x) e x cos(x) to both sides of this equation: e x cos(x) e x sin(x)+e x cos(x). Now divide both sides by to obtain the answer: e x cos(x) ex ( ) sin(x) + cos(x). WARNING: It is important that in the second step you do not choose u sin(x) and dv e x. If you do, you will end up undoing the first step.

8 Math 4Q 7. x 4 ( /4 x + + /4 ) x 4 ln x + + ln x + C 4 8. x x 4 ( / x + + / ) x ln x + + ln x + C 9. x(x + ) ( x + ) x + ln x ln x + + C 3. x (x + ) [Use A x, B x, C x+.] ( x + x + ) x + ln x + ln x + + C x

9 Math 4Q 3. x x 6 ( 5/8 x /8 ) x ln x ln x 4 + C 8 3. x +7 x (x + ) You should use partial fractions with A, B C. Solving for the A, B, C, x x x+ you will get A 5 4, B 7, C 5 4. Then the integral will be equal to ( 5 ) x + x + 4 x ln x + 7 ( ) + 5 ln x + + C x x(x + ) You should use partial fractions with A, Bx+C. Solving for A, B, C, x x + you will get A, B, C. Then the integral will be equal to ( ) ( ( )x + + x x + x x ) x + ln x ln x + + C Note: To do the second integral, use substitution with u x +. We don t need absolute value signs on the second term, because x + is always positive.

10 Math 4Q 34. e x + [Hint: This one is really tricky. Multiply by ex and then use substitution.] e x Following the hint you get: e x + e x e x (e x + ) Now let u e x, so that du e x. and you get: e x (e x + ) ex. Substitute this into the integral e x (e x + ) ex u(u + ) du ( u + ) du u + ln u ln u + + C Put back the x s and you get the answer: e x + ln ex ln e x + + C. Note: Since e x is always positive, you don t need the absolute value signs.

11 Math 4Q 35. +x arctan(x)+c [x tan(θ) θ arctan(x)] x arcsin( x )+C [x sin(θ) θ arcsin( x )] 37. x +4x +5 (x + ) + arctan(x + ) + C [x + tan(θ) θ arctan(x + )] x 38. +x [Hint: This one is tricky. You might have to add and subtract from the numerator.] x + ( ) x + ( x + x + ) x + ( ) x arctan(x)+c x +

12 Math 4Q 39. x r x x r + r (as r ) 4. x R x x R + + (as R ) R 4. e ln x x R e ln x x (ln x) R e (ln R) (as R ) [Use substitution with u ln x] 4. e x(ln x) R e x(ln x) R ln x e + + (as R ) ln R [Use substitution with u ln x]

13 Math 4Q 43. xe x Use integration by parts with u x and dv e x. R xe x ( xe x + Then ) e x R ( xe x e x) R as R. ( Re R e R) ( ) + Remark: Re R e R R e (as R ). R er 44. xe x Use substitution with u x and du x. Then R xe x R e x ( ) ( e R ) + as R. Remark: e R e R (as R ).

14 Math 4Q 45. arctan x x + [Hint: What is the derivative of arctan(x)?] Use substitution with u arctan(x) and du x +. R as R. arctan x x + (arctan x) ( π R Then (arctan R) (arctan ) ) ( ) π π Remarks:. arctan() π 4, because. lim R arctan R π, because ( ) π tan sin ( ) π 4 4 cos ( ) π 4 / /. ( ) π tan sin ( ) π cos ( ) π. This should be thought of as a limit, not actual equality. 46. f(x), where f(x) { x if x x if x ( + x x x) + ( x) + 3