Mixing
Integral using trig substitution check
$begingroup$
I am trying to integrate this:
$$int_0^2 frac{dt}{sqrt{4 + t^2}}$$
after letting $t = 2 tan {theta}$, I find that the equation reduces to:
$$int sec {theta}, d theta = ln | sec {theta} + tan {theta} |$$
and I find that the indefinite integral after replacing the thetas is is:
$$ln Bigg| frac{sqrt{t^2 + 4}}{2} + t ,Bigg| $$
when I take the definite integral I get:
$$ln big|, 2 + sqrt{2} ,big| $$ which is not what wolfram gets:
What am I doing wrong?
integration
edited Feb 3 at 5:38
El borito
664216
asked Feb 2 at 23:30
Jwan622Jwan622
2,38511632
$endgroup$
add a comment |
$begingroup$
I am trying to integrate this:
$$int_0^2 frac{dt}{sqrt{4 + t^2}}$$
after letting $t = 2 tan {theta}$, I find that the equation reduces to:
$$int sec {theta}, d theta = ln | sec {theta} + tan {theta} |$$
and I find that the indefinite integral after replacing the thetas is is:
$$ln Bigg| frac{sqrt{t^2 + 4}}{2} + t ,Bigg| $$
when I take the definite integral I get:
$$ln big|, 2 + sqrt{2} ,big| $$ which is not what wolfram gets:
What am I doing wrong?
integration
edited Feb 3 at 5:38
El borito
664216
asked Feb 2 at 23:30
Jwan622Jwan622
2,38511632
$endgroup$
add a comment |
$begingroup$
I am trying to integrate this:
$$int_0^2 frac{dt}{sqrt{4 + t^2}}$$
after letting $t = 2 tan {theta}$, I find that the equation reduces to:
$$int sec {theta}, d theta = ln | sec {theta} + tan {theta} |$$
and I find that the indefinite integral after replacing the thetas is is:
$$ln Bigg| frac{sqrt{t^2 + 4}}{2} + t ,Bigg| $$
when I take the definite integral I get:
$$ln big|, 2 + sqrt{2} ,big| $$ which is not what wolfram gets:
What am I doing wrong?
integration
edited Feb 3 at 5:38
El borito
664216
asked Feb 2 at 23:30
Jwan622Jwan622
2,38511632
$endgroup$
I am trying to integrate this:
$$int_0^2 frac{dt}{sqrt{4 + t^2}}$$
after letting $t = 2 tan {theta}$, I find that the equation reduces to:
$$int sec {theta}, d theta = ln | sec {theta} + tan {theta} |$$
and I find that the indefinite integral after replacing the thetas is is:
$$ln Bigg| frac{sqrt{t^2 + 4}}{2} + t ,Bigg| $$
when I take the definite integral I get:
$$ln big|, 2 + sqrt{2} ,big| $$ which is not what wolfram gets:
What am I doing wrong?
integration
integration
edited Feb 3 at 5:38
El borito
664216
asked Feb 2 at 23:30
Jwan622Jwan622
2,38511632
edited Feb 3 at 5:38
El borito
664216
asked Feb 2 at 23:30
Jwan622Jwan622
2,38511632
edited Feb 3 at 5:38
El borito
664216
edited Feb 3 at 5:38
El borito
664216
edited Feb 3 at 5:38
El borito
664216
664216
asked Feb 2 at 23:30
Jwan622Jwan622
2,38511632
asked Feb 2 at 23:30
Jwan622Jwan622
2,38511632
asked Feb 2 at 23:30
Jwan622Jwan622
2,38511632
2,38511632
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You made an algebra error when back substituting. It should be
$$
lnleft|frac{1}{2}sqrt{t^2+4} + frac{t}{2}right|
$$
This gives $ln(1+sqrt{2}) = sinh^{-1}(1)$.
answered Feb 2 at 23:34
eyeballfrogeyeballfrog
7,212633
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
You made an algebra error when back substituting. It should be
$$
lnleft|frac{1}{2}sqrt{t^2+4} + frac{t}{2}right|
$$
This gives $ln(1+sqrt{2}) = sinh^{-1}(1)$.
answered Feb 2 at 23:34
eyeballfrogeyeballfrog
7,212633
$endgroup$
add a comment |
$begingroup$
You made an algebra error when back substituting. It should be
$$
lnleft|frac{1}{2}sqrt{t^2+4} + frac{t}{2}right|
$$
This gives $ln(1+sqrt{2}) = sinh^{-1}(1)$.
answered Feb 2 at 23:34
eyeballfrogeyeballfrog
7,212633
$endgroup$
add a comment |
$begingroup$
You made an algebra error when back substituting. It should be
$$
lnleft|frac{1}{2}sqrt{t^2+4} + frac{t}{2}right|
$$
This gives $ln(1+sqrt{2}) = sinh^{-1}(1)$.
answered Feb 2 at 23:34
eyeballfrogeyeballfrog
7,212633
$endgroup$
You made an algebra error when back substituting. It should be
$$
lnleft|frac{1}{2}sqrt{t^2+4} + frac{t}{2}right|
$$
This gives $ln(1+sqrt{2}) = sinh^{-1}(1)$.
answered Feb 2 at 23:34
eyeballfrogeyeballfrog
7,212633
edited Feb 3 at 3:59
answered Feb 2 at 23:34
eyeballfrogeyeballfrog
7,212633
answered Feb 2 at 23:34
eyeballfrogeyeballfrog
7,212633
answered Feb 2 at 23:34
eyeballfrogeyeballfrog
7,212633
7,212633
add a comment |
add a comment |
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