Mixing
Evaluate the indefinite integral $int frac{1}{x^2} sinleft(frac{6}{x}right) cosleft(frac{6}{x}right) , dx $
$begingroup$
Evaluate the indefinite integral:
$$int frac{1}{x^2} sinleft(frac{6}{x}right) cosleft(frac{6}{x}right) , dx $$
(using substitution)
The answer is: $$frac {1}{24} cosleft(frac{12}{x}right) + C $$
Whereas I get a slightly different one.
Here's my solution:
$$u = frac{6}{x}$$
$$du = - frac{6}{x^2} cdot dx$$
$$-frac {1}{6} du = frac {1}{x^2} cdot dx$$
Making substitution:
$$ int -frac{1}{6} du sin (u) cos (u) $$
adding a new variable for substitution:
$$s = cos (u)$$
$$ds = -sin(u) du$$
Making substitution:
$$ frac {1}{6} int ds cdot s $$
Evaluating:
$$ frac{1}{6} cdot frac{s^2}{2} = frac{s^2}{12} = frac{cos^2(u)}{12} = frac{cos^2(frac{6}{x})}{12}$$
Then using a half angle formula for $cos^2$:
$$ frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} = frac{1}{24} + frac{1}{24} cdot cosleft(frac{12}{x}right) + C$$
As you can see I have an additional $frac{1}{24}$ in my answer... so what did I do wrong?
calculus integration indefinite-integrals
edited Jan 23 at 2:17
Larry
2,52831131
asked Apr 27 '15 at 4:13
dramadeurdramadeur
2512411
$endgroup$
add a comment |
$begingroup$
Evaluate the indefinite integral:
$$int frac{1}{x^2} sinleft(frac{6}{x}right) cosleft(frac{6}{x}right) , dx $$
(using substitution)
The answer is: $$frac {1}{24} cosleft(frac{12}{x}right) + C $$
Whereas I get a slightly different one.
Here's my solution:
$$u = frac{6}{x}$$
$$du = - frac{6}{x^2} cdot dx$$
$$-frac {1}{6} du = frac {1}{x^2} cdot dx$$
Making substitution:
$$ int -frac{1}{6} du sin (u) cos (u) $$
adding a new variable for substitution:
$$s = cos (u)$$
$$ds = -sin(u) du$$
Making substitution:
$$ frac {1}{6} int ds cdot s $$
Evaluating:
$$ frac{1}{6} cdot frac{s^2}{2} = frac{s^2}{12} = frac{cos^2(u)}{12} = frac{cos^2(frac{6}{x})}{12}$$
Then using a half angle formula for $cos^2$:
$$ frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} = frac{1}{24} + frac{1}{24} cdot cosleft(frac{12}{x}right) + C$$
As you can see I have an additional $frac{1}{24}$ in my answer... so what did I do wrong?
calculus integration indefinite-integrals
edited Jan 23 at 2:17
Larry
2,52831131
asked Apr 27 '15 at 4:13
dramadeurdramadeur
2512411
$endgroup$
5
$begingroup$
Did you notice the +C in the original answer there? The constant absorbs that $+1/24$!
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:18
$begingroup$
@Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C...
$endgroup$
– dramadeur
Apr 27 '15 at 4:20
1
$begingroup$
@dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + frac{1}{24}$ and then renaming $C'$ as $C$.
$endgroup$
– Travis
Apr 27 '15 at 4:23
$begingroup$
@Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:)
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:25
add a comment |
$begingroup$
Evaluate the indefinite integral:
$$int frac{1}{x^2} sinleft(frac{6}{x}right) cosleft(frac{6}{x}right) , dx $$
(using substitution)
The answer is: $$frac {1}{24} cosleft(frac{12}{x}right) + C $$
Whereas I get a slightly different one.
Here's my solution:
$$u = frac{6}{x}$$
$$du = - frac{6}{x^2} cdot dx$$
$$-frac {1}{6} du = frac {1}{x^2} cdot dx$$
Making substitution:
$$ int -frac{1}{6} du sin (u) cos (u) $$
adding a new variable for substitution:
$$s = cos (u)$$
$$ds = -sin(u) du$$
Making substitution:
$$ frac {1}{6} int ds cdot s $$
Evaluating:
$$ frac{1}{6} cdot frac{s^2}{2} = frac{s^2}{12} = frac{cos^2(u)}{12} = frac{cos^2(frac{6}{x})}{12}$$
Then using a half angle formula for $cos^2$:
$$ frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} = frac{1}{24} + frac{1}{24} cdot cosleft(frac{12}{x}right) + C$$
As you can see I have an additional $frac{1}{24}$ in my answer... so what did I do wrong?
calculus integration indefinite-integrals
edited Jan 23 at 2:17
Larry
2,52831131
asked Apr 27 '15 at 4:13
dramadeurdramadeur
2512411
$endgroup$
Evaluate the indefinite integral:
$$int frac{1}{x^2} sinleft(frac{6}{x}right) cosleft(frac{6}{x}right) , dx $$
(using substitution)
The answer is: $$frac {1}{24} cosleft(frac{12}{x}right) + C $$
Whereas I get a slightly different one.
Here's my solution:
$$u = frac{6}{x}$$
$$du = - frac{6}{x^2} cdot dx$$
$$-frac {1}{6} du = frac {1}{x^2} cdot dx$$
Making substitution:
$$ int -frac{1}{6} du sin (u) cos (u) $$
adding a new variable for substitution:
$$s = cos (u)$$
$$ds = -sin(u) du$$
Making substitution:
$$ frac {1}{6} int ds cdot s $$
Evaluating:
$$ frac{1}{6} cdot frac{s^2}{2} = frac{s^2}{12} = frac{cos^2(u)}{12} = frac{cos^2(frac{6}{x})}{12}$$
Then using a half angle formula for $cos^2$:
$$ frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} = frac{1}{24} + frac{1}{24} cdot cosleft(frac{12}{x}right) + C$$
As you can see I have an additional $frac{1}{24}$ in my answer... so what did I do wrong?
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Jan 23 at 2:17
Larry
2,52831131
asked Apr 27 '15 at 4:13
dramadeurdramadeur
2512411
edited Jan 23 at 2:17
Larry
2,52831131
asked Apr 27 '15 at 4:13
dramadeurdramadeur
2512411
edited Jan 23 at 2:17
Larry
2,52831131
edited Jan 23 at 2:17
Larry
2,52831131
edited Jan 23 at 2:17
Larry
2,52831131
2,52831131
asked Apr 27 '15 at 4:13
dramadeurdramadeur
2512411
asked Apr 27 '15 at 4:13
dramadeurdramadeur
2512411
asked Apr 27 '15 at 4:13
dramadeurdramadeur
2512411
2512411
5
$begingroup$
Did you notice the +C in the original answer there? The constant absorbs that $+1/24$!
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:18
$begingroup$
@Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C...
$endgroup$
– dramadeur
Apr 27 '15 at 4:20
1
$begingroup$
@dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + frac{1}{24}$ and then renaming $C'$ as $C$.
$endgroup$
– Travis
Apr 27 '15 at 4:23
$begingroup$
@Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:)
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:25
add a comment |
5
$begingroup$
Did you notice the +C in the original answer there? The constant absorbs that $+1/24$!
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:18
$begingroup$
@Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C...
$endgroup$
– dramadeur
Apr 27 '15 at 4:20
1
$begingroup$
@dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + frac{1}{24}$ and then renaming $C'$ as $C$.
$endgroup$
– Travis
Apr 27 '15 at 4:23
$begingroup$
@Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:)
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:25
5
5
$begingroup$
Did you notice the +C in the original answer there? The constant absorbs that $+1/24$!
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:18
$begingroup$
Did you notice the +C in the original answer there? The constant absorbs that $+1/24$!
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:18
$begingroup$
@Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C...
$endgroup$
– dramadeur
Apr 27 '15 at 4:20
$begingroup$
@Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C...
$endgroup$
– dramadeur
Apr 27 '15 at 4:20
1
1
$begingroup$
@dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + frac{1}{24}$ and then renaming $C'$ as $C$.
$endgroup$
– Travis
Apr 27 '15 at 4:23
$begingroup$
@dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + frac{1}{24}$ and then renaming $C'$ as $C$.
$endgroup$
– Travis
Apr 27 '15 at 4:23
$begingroup$
@Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:)
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:25
$begingroup$
@Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:)
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You missed the constant of integration all the way! It "absorbs" the $frac{1}{24}$!
Spot the difference:
Evaluating:
$displaystyle frac{1}{6} cdot frac{s^2}{2} +c= frac{s^2}{12} +c= frac{cos^2(u)}{12} +c= frac{cos^2(frac{6}{x})}{12}+c$, where c is the constant of integration.
Then using a half angle formula for $cos^2$:
$displaystyle frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} +c$ = $frac{1}{24} cdot cos(frac{12}{x}) + C$, where $C=c+frac{1}{24}$
answered Apr 27 '15 at 4:24
Jesse P FrancisJesse P Francis
56911656
$endgroup$
add a comment |
$begingroup$
You did everything right, but you're constant of integration "absorbs" the $frac{1}{24}$
$$frac{1}{24}+C=C$$
The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant!
If it helps, you could do something like:
$$frac{1}{24}+C=D$$
answered Apr 27 '15 at 4:33
bthmasbthmas
57439
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
You missed the constant of integration all the way! It "absorbs" the $frac{1}{24}$!
Spot the difference:
Evaluating:
$displaystyle frac{1}{6} cdot frac{s^2}{2} +c= frac{s^2}{12} +c= frac{cos^2(u)}{12} +c= frac{cos^2(frac{6}{x})}{12}+c$, where c is the constant of integration.
Then using a half angle formula for $cos^2$:
$displaystyle frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} +c$ = $frac{1}{24} cdot cos(frac{12}{x}) + C$, where $C=c+frac{1}{24}$
answered Apr 27 '15 at 4:24
Jesse P FrancisJesse P Francis
56911656
$endgroup$
add a comment |
$begingroup$
You missed the constant of integration all the way! It "absorbs" the $frac{1}{24}$!
Spot the difference:
Evaluating:
$displaystyle frac{1}{6} cdot frac{s^2}{2} +c= frac{s^2}{12} +c= frac{cos^2(u)}{12} +c= frac{cos^2(frac{6}{x})}{12}+c$, where c is the constant of integration.
Then using a half angle formula for $cos^2$:
$displaystyle frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} +c$ = $frac{1}{24} cdot cos(frac{12}{x}) + C$, where $C=c+frac{1}{24}$
answered Apr 27 '15 at 4:24
Jesse P FrancisJesse P Francis
56911656
$endgroup$
add a comment |
$begingroup$
You missed the constant of integration all the way! It "absorbs" the $frac{1}{24}$!
Spot the difference:
Evaluating:
$displaystyle frac{1}{6} cdot frac{s^2}{2} +c= frac{s^2}{12} +c= frac{cos^2(u)}{12} +c= frac{cos^2(frac{6}{x})}{12}+c$, where c is the constant of integration.
Then using a half angle formula for $cos^2$:
$displaystyle frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} +c$ = $frac{1}{24} cdot cos(frac{12}{x}) + C$, where $C=c+frac{1}{24}$
answered Apr 27 '15 at 4:24
Jesse P FrancisJesse P Francis
56911656
$endgroup$
You missed the constant of integration all the way! It "absorbs" the $frac{1}{24}$!
Spot the difference:
Evaluating:
$displaystyle frac{1}{6} cdot frac{s^2}{2} +c= frac{s^2}{12} +c= frac{cos^2(u)}{12} +c= frac{cos^2(frac{6}{x})}{12}+c$, where c is the constant of integration.
Then using a half angle formula for $cos^2$:
$displaystyle frac {frac{1}{2} cdot (1 + cos(frac{12}{x}))}{12} +c$ = $frac{1}{24} cdot cos(frac{12}{x}) + C$, where $C=c+frac{1}{24}$
answered Apr 27 '15 at 4:24
Jesse P FrancisJesse P Francis
56911656
answered Apr 27 '15 at 4:24
Jesse P FrancisJesse P Francis
56911656
answered Apr 27 '15 at 4:24
Jesse P FrancisJesse P Francis
56911656
answered Apr 27 '15 at 4:24
Jesse P FrancisJesse P Francis
56911656
56911656
add a comment |
add a comment |
$begingroup$
You did everything right, but you're constant of integration "absorbs" the $frac{1}{24}$
$$frac{1}{24}+C=C$$
The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant!
If it helps, you could do something like:
$$frac{1}{24}+C=D$$
answered Apr 27 '15 at 4:33
bthmasbthmas
57439
$endgroup$
add a comment |
$begingroup$
You did everything right, but you're constant of integration "absorbs" the $frac{1}{24}$
$$frac{1}{24}+C=C$$
The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant!
If it helps, you could do something like:
$$frac{1}{24}+C=D$$
answered Apr 27 '15 at 4:33
bthmasbthmas
57439
$endgroup$
add a comment |
$begingroup$
You did everything right, but you're constant of integration "absorbs" the $frac{1}{24}$
$$frac{1}{24}+C=C$$
The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant!
If it helps, you could do something like:
$$frac{1}{24}+C=D$$
answered Apr 27 '15 at 4:33
bthmasbthmas
57439
$endgroup$
You did everything right, but you're constant of integration "absorbs" the $frac{1}{24}$
$$frac{1}{24}+C=C$$
The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant!
If it helps, you could do something like:
$$frac{1}{24}+C=D$$
answered Apr 27 '15 at 4:33
bthmasbthmas
57439
answered Apr 27 '15 at 4:33
bthmasbthmas
57439
answered Apr 27 '15 at 4:33
bthmasbthmas
57439
answered Apr 27 '15 at 4:33
bthmasbthmas
57439
57439
add a comment |
add a comment |
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5
$begingroup$
Did you notice the +C in the original answer there? The constant absorbs that $+1/24$!
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:18
$begingroup$
@Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C...
$endgroup$
– dramadeur
Apr 27 '15 at 4:20
1
$begingroup$
@dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + frac{1}{24}$ and then renaming $C'$ as $C$.
$endgroup$
– Travis
Apr 27 '15 at 4:23
$begingroup$
@Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:)
$endgroup$
– Jesse P Francis
Apr 27 '15 at 4:25