Sarah Howe 

Calc II Spring '07 

Brewer 

Honor's Footnote 

 





 

 

int(1/((x+2)*(x+3)), x) 

ln(x+2)-ln(x+3) 

 

int(1/((x+1)*(x+5)), x) 

1/4*ln(x+1)-1/4*ln(x+5) 

int(1/((x+2)*(x-5)), x) 

-1/7*ln(x+2)+1/7*ln(x-5) 

int(1/(x+2)^2, x) 

-1/(x+2) 

int(1/(x+a)^2, x) 

-1/(x+a) 

int(1/((x+a)*(x+b)), x) 

 

Based on the pattern seen above int(1/((x+a)*(x+b)), x) = ln(x+b)/(-b+a)-ln(x+a)/(-b+a) 

 

diff([-ln(x+a)/(-b+a)+ln(x+b)/(-b+a)], x) 

[-1/((x+a)*(-b+a))+1/((x+b)*(-b+a))] 

 

 

Verification*of*this*hypothesis*follows; -1 

int(1/((x+a)*(x+b)), x) 

ln(x+b)/(-b+a)-ln(x+a)/(-b+a) 

diff(ln(x+b)/(-b+a)-ln(x+a)/(-b+a), x) 

1/((x+b)*(-b+a))-1/((x+a)*(-b+a)) 

 

 

The second family of integrals I will look at involves trigonometric functions. 

 

 

int(sin(x)*cos(2*x), x) 

-1/6*cos(3*x)+1/2*cos(x) 

int(sin(3*x)*cos(7*x), x) 

-1/20*cos(10*x)+1/8*cos(4*x) 

int(sin(8*x)*cos(3*x), x) 

-1/22*cos(11*x)-1/10*cos(5*x) 

 

 

Based on the patterns seen above int(sin(a*x)*cos(b*x), x) = -cos((a+b)*x)/(2*a+2*b)-cos((-b+a)*x)/(-2*b+2*a) 

diff(-1/2*cos((a+b)*x)/(a+b)-1/2*cos((-b+a)*x)/(-b+a), x) 

1/2*sin((a+b)*x)+1/2*sin((-b+a)*x) 

 


 Applying the addition and subtraction identities yields the following:

1/2*sin(ax+bx)+1/2*sin(ax-bx) = [1/2*`+`(sin(ax)*cos(bx), cos(ax)*sin(bx), sin(ax)*cos(bx), -cos(ax)*sin(bx))]
1/2*sin(ax+bx)+1/2*sin(ax-bx) = [1/2*`+`(sin(ax)*cos(bx), cos(ax)*sin(bx), sin(ax)*cos(bx), -cos(ax)*sin(bx))] 

 

 

 

1/2*sin(ax+bx)+1/2*sin(ax-bx) = sin(ax)*cos(bx) 

 

The next familiy of integrals includes the natural log function. 

 

int(ln(x), x) 

x*ln(x)-x 

int(x*ln(x), x) 

1/2*x^2*ln(x)-1/4*x^2 

int(x^2*ln(x), x) 

1/3*x^3*ln(x)-1/9*x^3 

int(x^3*ln(x), x) 

1/4*x^4*ln(x)-1/16*x^4 

int(x^7*ln(x), x) 

1/8*x^8*ln(x)-1/64*x^8 

 

Based*on*the*patterns*seen*above:
 

Verification*of*the*above*hypothesis*follows; -1 

 

diff(ln(x)*x^(n+1)/(n+1)-x^(n+1)/(n^2+2*n+1), x) 

x^(n+1)/(x*(n+1))+ln(x)*x^(n+1)/x-x^(n+1)*(n+1)/(x*(n^2+2*n+1)) 

simplify(%) 

ln(x)*x^n 

 


 

 

 

int(x*e^x, x) 

(-1+x*ln(e))*e^x/ln(e)^2 

int(x^2*e^x, x) 

(2-2*x*ln(e)+x^2*ln(e)^2)*e^x/ln(e)^3 

int(x^3*e^x, x) 

(-6+6*x*ln(e)-3*x^2*ln(e)^2+x^3*ln(e)^3)*e^x/ln(e)^4 

int(x^4*e^x, x) 

(24-24*x*ln(e)+12*x^2*ln(e)^2-4*x^3*ln(e)^3+x^4*ln(e)^4)*e^x/ln(e)^5 

int(x^5*e^x, x) 

(-120+120*x*ln(e)-60*x^2*ln(e)^2+20*x^3*ln(e)^3-5*x^4*ln(e)^4+x^5*ln(e)^5)*e^x/ln(e)^6 

int(x^6*e^x, x) 

(720-720*x*ln(e)+360*x^2*ln(e)^2-120*x^3*ln(e)^3+30*x^4*ln(e)^4-6*x^5*ln(e)^5+x^6*ln(e)^6)*e^x/ln(e)^7
(720-720*x*ln(e)+360*x^2*ln(e)^2-120*x^3*ln(e)^3+30*x^4*ln(e)^4-6*x^5*ln(e)^5+x^6*ln(e)^6)*e^x/ln(e)^7 

 

Based on the patterns above int(x^n*e^x, x) = x^n*e^x-n*int(x^(n-1)*e^x, x) 

Verification by mathematical induction: 



 

2) Showing the expression is also true for n+1 given that it is true for n. 

 

int(x^(n+1)*e^x, x) = x^(n+1)*e^x-(n+1)*int(x^n*e^x, x) 

 

diff([x^(n+1)*e^x-(n+1)*int(x^n*e^x, x)], x) = `+`((n+1)*x^n*e^x, x^(n+1)*e^x, -(n+1)*x^n*e^x)
diff([x^(n+1)*e^x-(n+1)*int(x^n*e^x, x)], x) = `+`((n+1)*x^n*e^x, x^(n+1)*e^x, -(n+1)*x^n*e^x)
diff([x^(n+1)*e^x-(n+1)*int(x^n*e^x, x)], x) = `+`((n+1)*x^n*e^x, x^(n+1)*e^x, -(n+1)*x^n*e^x)= x^(n+1)*e^x