VIEWS: 0 PAGES: 3 CATEGORY: Business POSTED ON: 2/23/2011 Public Domain
Integration Worksheet Math 132 Solutions (a) Use partial fractions: x+3 1 10 1 1 2 + 5x − 2 dx = dx + dx 3x 7 3x − 1 7 x+2 10 1 = ln(3x − 1) + ln(x + 2) + C. 21 7 (b) Use integration by parts with u = x, dv = e3x dx (which gives du = dx and 1 v = 3 e3x ): 1 1 1 x x 1 3x dx = e3x − e3x dx 0 e 3 0 3 0 1 1 x 1 = e3x − e3x 3 0 9 0 3 1 + 2e = . 9 √ (c) Use u substitution. Let u = x + 1 to get u2 = x + 1, 2udu = dx, x = u2 − 1. Also, x = 0 ⇒ u = 1 and x = 8 ⇒ u = 3. 8 √ u=3 x x + 1 dx = (u2 − 1)(u)(2udu) 0 u=1 3 2u3 2u5 =− + 3 5 1 1192 = . 15 1−cos(2t) (d) Use the “half-angle” formula from trigonometry to write sin2 (t) = 2 . 1 − cos(2t) sin2 (t) dt = dt 2 1 1 = t − sin(2t) + C. 2 4 (e) Here, use the fact that cos2 (t) = 1 − sin2 (t): cos3 (t) dt = cos(t)(1 − sin2 (t)) dt sin3 (t) = sin(t) − + C. 3 (f) Here, I recommend using the substitution x = sinh(t). Then, x2 + 1 = sin2 (t) + 1 = cosh2 (t) and dx = cosh(t)dt. (I prefer the hyperbolic trig substitution here to the trigonometric substitution x = tan(t) because, in the end, I think it’s harder to deal with sec3 (t) than cosh2 (t).) √ x2 + 1 dx = cosh2 (t) cosh(t) dt = cosh2 (t) dt e2x + 2 + e−2x = dt 4 e2t + 4t − e−2t = +C 8 To ﬁnish, and put things in terms of x again, use the fact that x = sinh(t) ⇒ √ t = ln(x + 1 + x2 ) and a little algebra: √ √ √ (x + 1 + x2 )2 + 4 ln(x + 1 + x2 ) − (x + 1 + x2 )−2 = +C √ √ 8 x 1 + x2 ln(x + 1 + x2 ) = + + C. 2 2 (g) Use integration by parts with u = arcsin(x) and dv = dx (then we have du = √ dx and v = x). 1−x2 1 1 1 x dx arcsin(x) dx = x arcsin(x) − √ 0 0 0 1 − x2 1 √ 1 = x arcsin(x) + 1 − x2 0 0 π = − 1. 2 2 (h) Here, we use parts with u = (ln(x))2 and dv = dx (then we have du = 2 ln(x) dx x and v = x). e e e 2 2 (ln(x)) dx = x(ln(x)) − 2 ln(x) dx 1 1 1 e e 2 = x(ln(x)) + 2x − 2x ln(x) 1 1 = e − 2. 1 (i) Here use the fact that ln x 2 = 1 ln(x): 2 √ 1 ln x dx = ln(x) dx 2 x x ln(x) =− + + C. 2 2 (j) Here you have some choices. You could use the substitution x = 2 cosh(t). This time, we’ll use the trigonometric substitution x = 2 sec(t). This leads √ to x2 − 4 = 2 tan(t) and dx = 2 sec(t) tan(t)dt. Let’s ignore the limits of √ 20 dx integration in √5 x2 √x2 −4 at ﬁrst. dx 2 sec(t) tan(t) dt √ = x2 x2 − 4 8 sec2 (t) tan(t) 1 = cos(t) dt 4 1 = sin(t) + C. 4 Probably the tidiest way to ﬁnish, is to use the fact that when x = 2 sec(t) = √ √ 1 2 5 ⇒ sin(t) = √5 and x = 2 sec(t) = 20 ⇒ sin(t) = √5 . Then, the deﬁnite integral √ √ x= 20 x= 20 dx 1 1 2 1 1 √ = sin(t) = √ −√ = √ . √ x= 5 x2 x2 − 4 4 √ x= 5 4 5 5 4 5 3