14-December-2021 Homework
Find the integrals. (You can use any method you learned from the course. Not all problems require integration by parts.)
Solutions.
%2BC#card=math&code=%5Cint%20x%5E3%20%5Cln%20x%20%5C%2C%20dx%20%3D%20%5Cfrac%7Bx%5E4%7D%7B16%7D%284%5Cln%20x%20-1%29%2BC&id=zubU7)
%20%5C%2C%20dx%20%3D%20%5Cfrac%7B2x%2B1%7D%7B2%7D(%5Cln%20%7C2x%2B1%7C%20-1)%2BC#card=math&code=%5Cint%20%5Cln%282x%2B1%29%20%5C%2C%20dx%20%3D%20%5Cfrac%7B2x%2B1%7D%7B2%7D%28%5Cln%20%7C2x%2B1%7C%20-1%29%2BC&id=pe4YB)
%20%5C%2C%20dx%20%3D%20%5Cint%20%5Ctan%5E%7B-1%7D(x)%5C%2C%20dx%20%3D%20x%5Carctan%20x%20-%20(1%2F2)%5Cln(1%2Bx%5E2)%2BC#card=math&code=%5Cint%20%5Carctan%28x%29%20%5C%2C%20dx%20%3D%20%5Cint%20%5Ctan%5E%7B-1%7D%28x%29%5C%2C%20dx%20%3D%20x%5Carctan%20x%20-%20%281%2F2%29%5Cln%281%2Bx%5E2%29%2BC&id=MqUSp)
%5C%2C%20dx%20%3D%20(1%2F4)(%5Csin%20(2x)%20-%202x%20%5Ccos(2x))%2BC#card=math&code=%5Cint%20x%20%5Csin%282x%29%5C%2C%20dx%20%3D%20%281%2F4%29%28%5Csin%20%282x%29%20-%202x%20%5Ccos%282x%29%29%2BC&id=jyebo)
(4x%5E3-6x%5E2%2B6x-3)%2BC#card=math&code=%5Cint%20x%5E3%20%5Cmathrm%7Be%7D%5E%7B2x%7D%5C%2C%20dx%20%3D%20%28e%5E%7B2x%7D%2F8%29%284x%5E3-6x%5E2%2B6x-3%29%2BC&id=VBce3)
e%5E%7B-x%5E2%7D%2BC#card=math&code=%5Cint%20x%20%5Cmathrm%7Be%7D%5E%7B-x%5E2%7D%5C%2C%20dx%20%3D%20%28-1%2F2%29e%5E%7B-x%5E2%7D%2BC&id=rwiWY)
%20%5C%2C%20dx%2C%20%5Cquad%20I_2%20%3D%20%5Cint%20%5Ccos(%5Cln%20x)%5C%2C%20dx#card=math&code=I_1%20%3D%20%5Cint%20%5Csin%20%28%5Cln%20x%29%20%5C%2C%20dx%2C%20%5Cquad%20I_2%20%3D%20%5Cint%20%5Ccos%28%5Cln%20x%29%5C%2C%20dx&id=CGMl1) (Hint: let
, or you can use integration by parts and simultaneously solve out these two integrals)
Solution:%5Cleft(%5Csin%20(%5Cln%20x)-%5Ccos%20(%5Cln%20x)%5Cright)%2C%5Cquad%20%20I_2%20%3D(x%2F2)%20%5Cleft(%5Csin(%5Cln%20x)%2B%5Ccos(%5Cln%20(x))%20%5Cright)#card=math&code=I_1%20%3D%20%28x%2F2%29%5Cleft%28%5Csin%20%28%5Cln%20x%29-%5Ccos%20%28%5Cln%20x%29%5Cright%29%2C%5Cquad%20%20I_2%20%3D%28x%2F2%29%20%5Cleft%28%5Csin%28%5Cln%20x%29%2B%5Ccos%28%5Cln%20%28x%29%29%20%5Cright%29&id=kSO7Z).
%5E2%5C%2C%20dx%20%3D%20x(%5Cln%5E2(x)-2%5Cln%20(x)%2B2)%2BC#card=math&code=%5Cint%20%28%5Cln%20x%29%5E2%5C%2C%20dx%20%3D%20x%28%5Cln%5E2%28x%29-2%5Cln%20%28x%29%2B2%29%2BC&id=vpdfC)
%20%5C%2C%20dx%20%3D%20%5Cint%20%5Csin%5E%7B-1%7D%20x%20%5C%2C%20dx%20%3D%20x%5Carcsin(x)%2B%5Csqrt%7B1-x%5E2%7D%2BC#card=math&code=%5Cint%20%5Carcsin%28x%29%20%5C%2C%20dx%20%3D%20%5Cint%20%5Csin%5E%7B-1%7D%20x%20%5C%2C%20dx%20%3D%20x%5Carcsin%28x%29%2B%5Csqrt%7B1-x%5E2%7D%2BC&id=wUbCx)
%5C%2C%20dx%20%3D%20%5Cint%20x%20%5Ctan%5E%7B-1%7D%20(x)%20%5C%2C%20dx%20%3D%20(1%2F2)%5Cleft(%20x%20%5Carctan(x)%20-x%20%2B%20%5Carctan(x)%5Cright)%20%2BC#card=math&code=%5Cint%20x%20%5Carctan%28x%29%5C%2C%20dx%20%3D%20%5Cint%20x%20%5Ctan%5E%7B-1%7D%20%28x%29%20%5C%2C%20dx%20%3D%20%281%2F2%29%5Cleft%28%20x%20%5Carctan%28x%29%20-x%20%2B%20%5Carctan%28x%29%5Cright%29%20%2BC&id=YGt58)
%5E%7B5%2F2%7D%7D%7B5%7D%2B(2x-3)%5E%7B3%2F2%7D%2BC#card=math&code=%5Cint%202x%5Csqrt%7B2x-3%7D%5C%2C%20dx%20%3D%20%5Cfrac%7B%282x-3%29%5E%7B5%2F2%7D%7D%7B5%7D%2B%282x-3%29%5E%7B3%2F2%7D%2BC&id=MriJm)
Derive the following formulas using the technique of integration by parts. Assume that is a positive integer. These formulas are called reduction formulas because the exponent in the
term has been reduced by one in each case. The second integral is simpler than the original integral.
.
Solution:#card=math&code=In%20%3D%20x%5En%20e%5Ex%20-%20n%20I%7Bn-1%7D%5C%2C%20%28n%20%5Cgeq%201%29&id=hQxIJ). (Note that
e%5Ex%2BC%2C%20%20%5Cquad%20I_0%20%3D%20%5Cint%20e%5Ex%20%5C%2C%20dx%20%3D%20e%5Ex%20%2BC#card=math&code=I_1%20%3D%20%5Cint%20xe%5Ex%20%5C%2C%20dx%20%3D%28x-1%29e%5Ex%2BC%2C%20%20%5Cquad%20I_0%20%3D%20%5Cint%20e%5Ex%20%5C%2C%20dx%20%3D%20e%5Ex%20%2BC&id=Q6zsp).)
%5C%2C%20dx#card=math&code=In%3D%5Cint%20x%5En%20%5Ccos%20%28x%29%5C%2C%20dx&id=eo6Bu)
Solution:I%7Bn-2%7D#card=math&code=In%20%20%3D%20x%5En%20%5Csin%20x%20%2B%20nx%5E%7Bn-1%7D%5Ccos%20x%20-%20n%28n-1%29I%7Bn-2%7D&id=WoDJk). Here we have to apply the integration by parts twice to get the reduction formula.
(Note that%5C%2C%20dx%20%3D%20%5Csin%20(x)%2BC%2C%20%5Cquad%20I_1%20%3D%20%5Cint%20x%5Ccos(x)%5C%2C%20dx%3D%20x%5Csin(x)%2B%5Ccos(x)%2BC#card=math&code=I_0%20%3D%20%5Cint%20%5Ccos%20%28x%29%5C%2C%20dx%20%3D%20%5Csin%20%28x%29%2BC%2C%20%5Cquad%20I_1%20%3D%20%5Cint%20x%5Ccos%28x%29%5C%2C%20dx%3D%20x%5Csin%28x%29%2B%5Ccos%28x%29%2BC&id=IL7mx).)
%5C%2C%20dx#card=math&code=I_n%20%3D%20%5Cint%20%5Ctan%5En%28x%29%5C%2C%20dx&id=dsBpK) (Hint:
, and hence
%20)-%20dx#card=math&code=%5Ctan%5E2%20x%5C%2C%20dx%20%3D%20d%28%5Ctan%28x%29%20%29-%20dx&id=pER08).)
Solution: For, using the hint, we have
%5C%2C%20d(%5Ctan(x))%20-%20I_%7Bn-2%7D%20%5C%5C%0A%26%3D-I_%7Bn-2%7D%2B%5Ctan%5E%7Bn-1%7D(x)%20-%20%5Cint%20%5Ctan(x)%20(n-2)%20%5Ctan%5E%7Bn-3%7D(x)%20d(%5Ctan(x))%5C%5C%0A%26%3D-I_%7Bn-2%7D%2B%5Ctan%5E%7Bn-1%7D(x)-(n-2)%5Cint%20%5Ctan%5E%7Bn-2%7D(x)%5C%2C%20d(%5Ctan(x))%5C%5C%0A%26%3D%20-I_%7Bn-2%7D%2B%5Ctan%5E%7Bn-1%7D(x)-%5Cfrac%7Bn-2%7D%7Bn-1%7D%5Ctan%5E%7Bn-1%7D(x)%5C%5C%0A%26%3D-I_%7Bn-2%7D%2B%5Cfrac%7B1%7D%7Bn-1%7D%5Ctan%5E%7Bn-1%7D(x).%20%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0AI_n%20%26%3D%20%5Cint%20%5Ctan%5E%7Bn-2%7D%28x%29%5C%2C%20d%28%5Ctan%28x%29%29%20-%20I_%7Bn-2%7D%20%5C%5C%0A%26%3D-I_%7Bn-2%7D%2B%5Ctan%5E%7Bn-1%7D%28x%29%20-%20%5Cint%20%5Ctan%28x%29%20%28n-2%29%20%5Ctan%5E%7Bn-3%7D%28x%29%20d%28%5Ctan%28x%29%29%5C%5C%0A%26%3D-I_%7Bn-2%7D%2B%5Ctan%5E%7Bn-1%7D%28x%29-%28n-2%29%5Cint%20%5Ctan%5E%7Bn-2%7D%28x%29%5C%2C%20d%28%5Ctan%28x%29%29%5C%5C%0A%26%3D%20-I_%7Bn-2%7D%2B%5Ctan%5E%7Bn-1%7D%28x%29-%5Cfrac%7Bn-2%7D%7Bn-1%7D%5Ctan%5E%7Bn-1%7D%28x%29%5C%5C%0A%26%3D-I_%7Bn-2%7D%2B%5Cfrac%7B1%7D%7Bn-1%7D%5Ctan%5E%7Bn-1%7D%28x%29.%20%0A%5Cend%7Baligned%7D%0A&id=mQuJ2)<br /> Note that %20%5C%2C%20dx%20%3D%20-%5Cln(%5Ccos(x))%2BC#card=math&code=I_0%20%3D%20x%2BC%2C%20I_1%20%3D%20%5Cint%20%5Ctan%20%28x%29%20%5C%2C%20dx%20%3D%20-%5Cln%28%5Ccos%28x%29%29%2BC&id=AWjZh).