Question

4.(15 pts) Let a be a positive odd integer. Find the value of the following integral in terms of a using substitution with u = = sin(ax). jë sinº (ax)cosº (ax) dx 5. (10 pts) Let a, b, and cbe positive integers. Evaluate the following limit. If you use 2 Hospital's Rule, be sure to give indeterminare pe and mention when you inoke it. In (a+beck lim 06 x

Answer 1

4. Given u = sin(ax) = du = a cos(ax) dx Limits : x = 0= u = sin(ax) = sin(0) = 0 TT ал = & x=-=u= sin(ax) = sin 2 +1 (since a is odd) 2 2 sinº (ax )cosº (ax) dx 2 sin’ (ax )cos? (ax) cos(ax) dx x=0 x=0 2 x=0 sinº (ax)(1 – sinº (ax)) cos(ax) dx u" (1 (1–u?)du u– u du 1 +1 1111 =- u=0 a u=0 a

+1 6 n+1 1 u и u = using ſu"du = a 4 6 n+1 u=0 6 1 / (+1)* (+1) 1 1 = a 4 6 a4 6 12a

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In(a+be") In(a+be«») 5. lim x>0 x (by substituting limit directly, we can have indeterminate form 1 d In(a+be) (a+be) = lim a+bex dx .. lim X-> X X 1 (using L - Hospitals rule, differentiating numerator d and denominator, using formula In x dx 1 == X

= bcecx =lim (0+bcec ) = lim x+ a + becx x= a + becx bce* e bc bc = lim lim x+ a + becx CX bc = =C x>0 a 0+ b + b +6 - еех e e