Got two problems today:
Evaluate the integral:
1. (1)/((x-3)(x+3))dx
2. 92dx/(x^2(sqrt(x^2+64))) using the substitution x=8tan(theta)
Thanks!
Evaluate the integral:
1. (1)/((x-3)(x+3))dx
2. 92dx/(x^2(sqrt(x^2+64))) using the substitution x=8tan(theta)
Thanks!
-
Partial fractions:
1/(x^2-9)= a/(x+3) + b/(x-3)
1=ax-3a+bx+3b = x(a+b) + 1(3b-3a)
system: a+b=0; 3b-3a=1
We get a=-b; 3b-3(-b)=1; 6b=1; b=1/6 and a=-1/6
So we have (1/6)/(x-3) - (1/6)/(x+3)
Integrate: (1/6)[ln(x-3)-ln(x+3)]+C
Simplify (1/6)ln|(x-3)/(x+3)|+C
2] 92*INTEGRAL[dx/(x^2√(x^2+64))]
x=8tanu; dx=8(secu)^2 du
92*INTEGRAL[8(secu)^2/ [(tanu)^2*8secu] du]
92*INTEGRAL[cscu du]
92ln(cscu-cotu)+C
Now draw a right triangle with angle u
Since x=8tanu, tanu=x/8 and therefore
side opposite u=x; side adjacent u=8 and hypotenuse=√(x^2+64)
cscu=hyp/opp = √(x^2+64)/x and cotu=adj/opp= 8/x
Plug it all in and get 92ln[(√(x^2+64) - 8)/x] + C
1/(x^2-9)= a/(x+3) + b/(x-3)
1=ax-3a+bx+3b = x(a+b) + 1(3b-3a)
system: a+b=0; 3b-3a=1
We get a=-b; 3b-3(-b)=1; 6b=1; b=1/6 and a=-1/6
So we have (1/6)/(x-3) - (1/6)/(x+3)
Integrate: (1/6)[ln(x-3)-ln(x+3)]+C
Simplify (1/6)ln|(x-3)/(x+3)|+C
2] 92*INTEGRAL[dx/(x^2√(x^2+64))]
x=8tanu; dx=8(secu)^2 du
92*INTEGRAL[8(secu)^2/ [(tanu)^2*8secu] du]
92*INTEGRAL[cscu du]
92ln(cscu-cotu)+C
Now draw a right triangle with angle u
Since x=8tanu, tanu=x/8 and therefore
side opposite u=x; side adjacent u=8 and hypotenuse=√(x^2+64)
cscu=hyp/opp = √(x^2+64)/x and cotu=adj/opp= 8/x
Plug it all in and get 92ln[(√(x^2+64) - 8)/x] + C