4181H Problem Set 10 Selected Solutions. Chapter 15. = lim. (by L Hôpital s Rule) (0) = lim. x 2, = lim x 3 ( sin x)x + cos x cos x = lim
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1 48H Problem Set 0 Selected Solutions Chapter 5 # 3(a From the definition of the derivative we have sin f f( f(0 (0 sin = cos (by L Hôpital s Rule = sin (by L Hôpital s Rule = 0 (b From the definition of the derivative we have ( f f ( f (0 (0 0 0 Now f (0 = 0 by part (a, and for 0, f ( = by the quotient rule So ( becomes # 9(a We have ( (cos sin f (0 0 (cos sin, 0 (cos sin 3 ( sin + cos cos 0 3 sin (algebra 0 3 = 3 (by L Hôpital s Rule sin( + y sin( cos(y + cos( sin(y tan( + y = = cos( + y cos( cos(y sin( sin(y ; Now divide numerator and denominator by cos( cos(y: tan( + y = tan( + tan(y tan( tan(y This formula holds for any, y provided that, y, + y nπ/ for any odd integer n (these are the points where tan is undefined (b Let u, v be arbitrary, and let = arctan(u, y = arctan(v in (a; note that, y ( π/, π/ so that, y are acceptable in (, but we have to avoid u, v such that + y = ±π/ What eactly does this imply about u, v? Since, y ( π/, π/, + y = π/ implies that, y > 0 (so and y are complementary acute angles and thus uv = Similarly + y = π/ implies that, y < 0 (so and y are complementary acute angles and again uv = Conversely, if uv =, then u, v have the same sign If say u, v > 0, then from u = tan, v = tan y, and uv = we get tan tan y =, or tan = cot y So, y are complementary acute angles Similarly if u, v < 0, uv = implies that and y are complementary acute angles Thus
2 + y = ±π/ uv = Since tan( = u, tan(y = v: tan(arctan(u + arctan(v = u + v uv Thus if arctan(u + arctan(v ( π/, π/: If arctan(u + arctan(v > π/: If arctan(u + arctan(v < π/: arctan(u + arctan(v = arctan arctan(u + arctan(v = arctan arctan(u + arctan(v = arctan ( u + v uv ( u + v + π uv ( u + v π uv These formulas are based on the fact that tan( = u implies that arctan(u = + kπ for some integer k k can be determined uniquely from the requirement that π/ < + kπ < π/ # 6 From sin + cos = we get tan + = sec, so cos = sec = + tan Evaluating at arctan( we find cos (arctan( = + Since arctan( ( π/, π/, we know that cos(arctan( is positive, so Since sin = tan cos, we have cos(arctan( = + sin(arctan( = tan(arctan( cos(arctan( = + # 7 We have, from the addition formulas for the sine and the cosine: ( u ( u ( sin(u = sin cos, cos(u = cos u ( sin u substituting u/ = arctan(, we find, from #6, and sin(u = + + = + ( ( cos(u = = + + +
3 3 Chapter 9 # 3(iv Take f( =, g ( = sin, so that we may take g( = cos : thus sin d = ( cos ( cos d = cos + cos d; now use parts again, in the last integral, with f( = and g ( = cos, so g( = sin, and so cos d = sin sin d and therefore = sin + cos, sin d = cos + ( sin + cos = ( cos + sin (vii Take f( = sec and g ( = sec, so g( = tan : sec 3 d = sec tan sec tan tan d = sec tan sec tan d = sec tan sec 3 d + sec d = sec tan + log(sec + tan sec 3 d, hence sec 3 d = sec tan + log(sec + tan (viii Take f( = cos(log, g ( =, so g( = : cos(log d = cos(log ( sin(log d = cos(log + sin(log d Now do the last integral by parts, using f( = sin(log, g ( =, g( = : sin(log d = sin(log (cos(log d = sin(log cos(log d Substituting this into the previous equation gives cos(log d = cos(log + sin(log cos(log d,
4 4 so cos(log d = ( cos(log + sin(log # 4(ii We use = tan u, d = sec u du; then + = tan u + = sec u At this point we have to worry about the sign of the secant The substitution we are using is (strictly speaking u = arctan ; so the possible values for u are in the range π < u < π ; for such u, sec u is positive, so sec u = sec u Therefore d sec + = u du = sec u sec u du = log(sec u + tan u We finish the computation by substituting u = arctan ; we have tan(arctan = ; to evaluate sec(arctan use sec u = tan u +, so that sec(arctan = + Hence d + = log( + + (viii Substitute = sin u (again the real substitution is u = arcsin : cos d = u cos u du = cos u du Again we have used the fact that π u π to say that cos u = cos u Net we use the trig identity cos u = ( + cos u/: + cos u cos u du = du = sin u u + = sin u cos u u + 4 Finally, we substitute u = arcsin to obtain d = arcsin + # 5 (iii The difficulty here is the different radicals in the denominator, and the solution is to find a substitution that resolves both of them, namely = u 6 : d 6u = du u 3 + u u 3 du = 6 u + = 6 u u + u + du = 6( u3 3 u + u log(u + = log( 6 + (v There s no obvious simplification, so we try u = tan (based on the general rule: if all else fails, substitute for the most unpleasant part of the integral or = arctan u, d = du/( + u : d + tan = + u du + u
5 5 Epand into partial fractions: ( + u( + u = a + u + bu + c + u ; we find a = /5, b = /5, c = /5, so that + u du + u = ( du 5 + u = 5 u du + u + du + u ( log( + u log( + u + arctan u, so that d + tan = ( log( + tan 5 log( + tan + = (log( + tan + log(cos + 5 (i Try = u ; then d = u du, so we get u d = u du u ( + u u = u u du ( + u = u du u Now let u = sin v: + sin v = sin v cos v dv cos v = sin v + sin v dv = cos v + cos v dv = cos v + v sin v cos v = u + arcsin u u u = + arcsin (If you start with the substitution u = instead, you obtain the answer arcsin ; why are these the same?
Substituting this into the differential equation and solving for u, we have. + u2. du = 3x + c. ( )
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