Solve the following eq [closed]
$begingroup$
Solve following integral $int ze^{2z} sin z dz$.
I am unable to solve this problem. Can anyone help me in solving it.
integration
edited Dec 25 '18 at 13:14
t.ysn
1397
asked Dec 25 '18 at 11:38
ABCDABCD
93
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closed as off-topic by mrtaurho, Namaste, Eevee Trainer, Carl Schildkraut, Brian Borchers Dec 26 '18 at 1:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Namaste, Eevee Trainer, Carl Schildkraut, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Solve following integral $int ze^{2z} sin z dz$.
I am unable to solve this problem. Can anyone help me in solving it.
integration
edited Dec 25 '18 at 13:14
t.ysn
1397
asked Dec 25 '18 at 11:38
ABCDABCD
93
$endgroup$
closed as off-topic by mrtaurho, Namaste, Eevee Trainer, Carl Schildkraut, Brian Borchers Dec 26 '18 at 1:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Namaste, Eevee Trainer, Carl Schildkraut, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.
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Use integration by parts two times
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– Dr. Sonnhard Graubner
Dec 25 '18 at 11:39
3
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Well what have you tried? Please show us your attempts, in MathJax, even if they are incorrect or don't lead anywhere. Perhaps you should start with integrating $zsin z$ first.
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– TheSimpliFire
Dec 25 '18 at 11:39
add a comment |
$begingroup$
Solve following integral $int ze^{2z} sin z dz$.
I am unable to solve this problem. Can anyone help me in solving it.
integration
edited Dec 25 '18 at 13:14
t.ysn
1397
asked Dec 25 '18 at 11:38
ABCDABCD
93
$endgroup$
Solve following integral $int ze^{2z} sin z dz$.
I am unable to solve this problem. Can anyone help me in solving it.
integration
integration
edited Dec 25 '18 at 13:14
t.ysn
1397
asked Dec 25 '18 at 11:38
ABCDABCD
93
edited Dec 25 '18 at 13:14
t.ysn
1397
asked Dec 25 '18 at 11:38
ABCDABCD
93
edited Dec 25 '18 at 13:14
t.ysn
1397
edited Dec 25 '18 at 13:14
t.ysn
1397
edited Dec 25 '18 at 13:14
t.ysn
1397
1397
asked Dec 25 '18 at 11:38
ABCDABCD
93
asked Dec 25 '18 at 11:38
ABCDABCD
93
asked Dec 25 '18 at 11:38
ABCDABCD
93
93
closed as off-topic by mrtaurho, Namaste, Eevee Trainer, Carl Schildkraut, Brian Borchers Dec 26 '18 at 1:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Namaste, Eevee Trainer, Carl Schildkraut, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, Namaste, Eevee Trainer, Carl Schildkraut, Brian Borchers Dec 26 '18 at 1:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Namaste, Eevee Trainer, Carl Schildkraut, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Use integration by parts two times
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 11:39
3
$begingroup$
Well what have you tried? Please show us your attempts, in MathJax, even if they are incorrect or don't lead anywhere. Perhaps you should start with integrating $zsin z$ first.
$endgroup$
– TheSimpliFire
Dec 25 '18 at 11:39
add a comment |
$begingroup$
Use integration by parts two times
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 11:39
3
$begingroup$
Well what have you tried? Please show us your attempts, in MathJax, even if they are incorrect or don't lead anywhere. Perhaps you should start with integrating $zsin z$ first.
$endgroup$
– TheSimpliFire
Dec 25 '18 at 11:39
$begingroup$
Use integration by parts two times
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 11:39
$begingroup$
Use integration by parts two times
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 11:39
3
3
$begingroup$
Well what have you tried? Please show us your attempts, in MathJax, even if they are incorrect or don't lead anywhere. Perhaps you should start with integrating $zsin z$ first.
$endgroup$
– TheSimpliFire
Dec 25 '18 at 11:39
$begingroup$
Well what have you tried? Please show us your attempts, in MathJax, even if they are incorrect or don't lead anywhere. Perhaps you should start with integrating $zsin z$ first.
$endgroup$
– TheSimpliFire
Dec 25 '18 at 11:39
add a comment |
2 Answers
2
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oldest
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$begingroup$
Integrate by parts: $int fg' = fg - int f'g$.
$f=x rightarrow f'=1$ and $g=e^{2x} sin{x} rightarrow g'= frac{2e^{2x} sin{x}-e^{2x} cos{x}}{5}$
So we have:
${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Now solving: ${displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Apply linearity: $=class{steps-node}{cssId{steps-node-1}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-2}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$.
Now solving:${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$, We will integrate by parts twice in a row, so we have:
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-{displaystyleint}dfrac{mathrm{e}^{2x}cosleft(xright)}{2},mathrm{d}x$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{4},mathrm{d}xright)$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}+class{steps-node}{cssId{steps-node-4}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}xright)$$
The integral ${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$ appears again on the right side of the equation, we can solve for it:
$$=dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5}$$
Now solving ${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$ same as above, we have:
$${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{2},mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}cosleft(xright)}{4},mathrm{d}xright)=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}+class{steps-node}{cssId{steps-node-6}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}xright)=dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{5}$$
Plug in solved integrals:
$$class{steps-node}{cssId{steps-node-7}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-8}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}-dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
and:
$$dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
The problem is solved:
$${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}+C=dfrac{mathrm{e}^{2x}left(left(10x-3right)sinleft(xright)+left(4-5xright)cosleft(xright)right)}{25}+C$$
answered Dec 25 '18 at 13:08
t.ysnt.ysn
1397
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add a comment |
$begingroup$
Note that
$$ int ze^{2z}sin z dz = operatorname{Im}left{int ze^{(2+i)z} dz right} $$
Using integration by parts
begin{align}
int z e^{(2+i)z} dz &= frac{1}{2+i}ze^{(2+i)z}- frac{1}{2+i}int e^{(2+i)z}dz \
&= frac{1}{2+i}ze^{(2+i)z} - frac{1}{(2+i)^2}e^{(2+i)z} + C \
&= frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z} + C
end{align}
Taking the imaginary part of the above, we obtain
$$ operatorname{Re}left{ frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z}right} = frac25 ze^{2z}sin z - frac15 ze^{2z}cos z - frac{3}{25}e^{2z}sin z + frac{4}{25}e^{2z}cos z $$
answered Dec 25 '18 at 13:29
DylanDylan
13.7k31027
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Integrate by parts: $int fg' = fg - int f'g$.
$f=x rightarrow f'=1$ and $g=e^{2x} sin{x} rightarrow g'= frac{2e^{2x} sin{x}-e^{2x} cos{x}}{5}$
So we have:
${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Now solving: ${displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Apply linearity: $=class{steps-node}{cssId{steps-node-1}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-2}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$.
Now solving:${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$, We will integrate by parts twice in a row, so we have:
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-{displaystyleint}dfrac{mathrm{e}^{2x}cosleft(xright)}{2},mathrm{d}x$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{4},mathrm{d}xright)$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}+class{steps-node}{cssId{steps-node-4}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}xright)$$
The integral ${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$ appears again on the right side of the equation, we can solve for it:
$$=dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5}$$
Now solving ${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$ same as above, we have:
$${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{2},mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}cosleft(xright)}{4},mathrm{d}xright)=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}+class{steps-node}{cssId{steps-node-6}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}xright)=dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{5}$$
Plug in solved integrals:
$$class{steps-node}{cssId{steps-node-7}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-8}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}-dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
and:
$$dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
The problem is solved:
$${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}+C=dfrac{mathrm{e}^{2x}left(left(10x-3right)sinleft(xright)+left(4-5xright)cosleft(xright)right)}{25}+C$$
answered Dec 25 '18 at 13:08
t.ysnt.ysn
1397
$endgroup$
add a comment |
$begingroup$
Integrate by parts: $int fg' = fg - int f'g$.
$f=x rightarrow f'=1$ and $g=e^{2x} sin{x} rightarrow g'= frac{2e^{2x} sin{x}-e^{2x} cos{x}}{5}$
So we have:
${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Now solving: ${displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Apply linearity: $=class{steps-node}{cssId{steps-node-1}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-2}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$.
Now solving:${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$, We will integrate by parts twice in a row, so we have:
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-{displaystyleint}dfrac{mathrm{e}^{2x}cosleft(xright)}{2},mathrm{d}x$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{4},mathrm{d}xright)$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}+class{steps-node}{cssId{steps-node-4}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}xright)$$
The integral ${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$ appears again on the right side of the equation, we can solve for it:
$$=dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5}$$
Now solving ${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$ same as above, we have:
$${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{2},mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}cosleft(xright)}{4},mathrm{d}xright)=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}+class{steps-node}{cssId{steps-node-6}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}xright)=dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{5}$$
Plug in solved integrals:
$$class{steps-node}{cssId{steps-node-7}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-8}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}-dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
and:
$$dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
The problem is solved:
$${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}+C=dfrac{mathrm{e}^{2x}left(left(10x-3right)sinleft(xright)+left(4-5xright)cosleft(xright)right)}{25}+C$$
answered Dec 25 '18 at 13:08
t.ysnt.ysn
1397
$endgroup$
add a comment |
$begingroup$
Integrate by parts: $int fg' = fg - int f'g$.
$f=x rightarrow f'=1$ and $g=e^{2x} sin{x} rightarrow g'= frac{2e^{2x} sin{x}-e^{2x} cos{x}}{5}$
So we have:
${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Now solving: ${displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Apply linearity: $=class{steps-node}{cssId{steps-node-1}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-2}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$.
Now solving:${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$, We will integrate by parts twice in a row, so we have:
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-{displaystyleint}dfrac{mathrm{e}^{2x}cosleft(xright)}{2},mathrm{d}x$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{4},mathrm{d}xright)$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}+class{steps-node}{cssId{steps-node-4}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}xright)$$
The integral ${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$ appears again on the right side of the equation, we can solve for it:
$$=dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5}$$
Now solving ${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$ same as above, we have:
$${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{2},mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}cosleft(xright)}{4},mathrm{d}xright)=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}+class{steps-node}{cssId{steps-node-6}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}xright)=dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{5}$$
Plug in solved integrals:
$$class{steps-node}{cssId{steps-node-7}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-8}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}-dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
and:
$$dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
The problem is solved:
$${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}+C=dfrac{mathrm{e}^{2x}left(left(10x-3right)sinleft(xright)+left(4-5xright)cosleft(xright)right)}{25}+C$$
answered Dec 25 '18 at 13:08
t.ysnt.ysn
1397
$endgroup$
Integrate by parts: $int fg' = fg - int f'g$.
$f=x rightarrow f'=1$ and $g=e^{2x} sin{x} rightarrow g'= frac{2e^{2x} sin{x}-e^{2x} cos{x}}{5}$
So we have:
${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Now solving: ${displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x$
Apply linearity: $=class{steps-node}{cssId{steps-node-1}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-2}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$.
Now solving:${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$, We will integrate by parts twice in a row, so we have:
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-{displaystyleint}dfrac{mathrm{e}^{2x}cosleft(xright)}{2},mathrm{d}x$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{4},mathrm{d}xright)$$
$$=dfrac{mathrm{e}^{2x}sinleft(xright)}{2}-left(dfrac{mathrm{e}^{2x}cosleft(xright)}{4}+class{steps-node}{cssId{steps-node-4}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}xright)$$
The integral ${displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x$ appears again on the right side of the equation, we can solve for it:
$$=dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5}$$
Now solving ${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x$ same as above, we have:
$${displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-{displaystyleint}-dfrac{mathrm{e}^{2x}sinleft(xright)}{2},mathrm{d}x=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}-{displaystyleint}-dfrac{mathrm{e}^{2x}cosleft(xright)}{4},mathrm{d}xright)=dfrac{mathrm{e}^{2x}cosleft(xright)}{2}-left(-dfrac{mathrm{e}^{2x}sinleft(xright)}{4}+class{steps-node}{cssId{steps-node-6}{dfrac{1}{4}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}xright)=dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{5}$$
Plug in solved integrals:
$$class{steps-node}{cssId{steps-node-7}{dfrac{2}{5}}}{displaystyleint}mathrm{e}^{2x}sinleft(xright),mathrm{d}x-class{steps-node}{cssId{steps-node-8}{dfrac{1}{5}}}{displaystyleint}mathrm{e}^{2x}cosleft(xright),mathrm{d}x=dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}-dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
and:
$$dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-{displaystyleint}dfrac{2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)}{5},mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}$$
The problem is solved:
$${displaystyleint}xmathrm{e}^{2x}sinleft(xright),mathrm{d}x=dfrac{xleft(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{5}-dfrac{2left(2mathrm{e}^{2x}sinleft(xright)-mathrm{e}^{2x}cosleft(xright)right)}{25}+dfrac{mathrm{e}^{2x}sinleft(xright)+2mathrm{e}^{2x}cosleft(xright)}{25}+C=dfrac{mathrm{e}^{2x}left(left(10x-3right)sinleft(xright)+left(4-5xright)cosleft(xright)right)}{25}+C$$
answered Dec 25 '18 at 13:08
t.ysnt.ysn
1397
answered Dec 25 '18 at 13:08
t.ysnt.ysn
1397
answered Dec 25 '18 at 13:08
t.ysnt.ysn
1397
answered Dec 25 '18 at 13:08
t.ysnt.ysn
1397
1397
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$begingroup$
Note that
$$ int ze^{2z}sin z dz = operatorname{Im}left{int ze^{(2+i)z} dz right} $$
Using integration by parts
begin{align}
int z e^{(2+i)z} dz &= frac{1}{2+i}ze^{(2+i)z}- frac{1}{2+i}int e^{(2+i)z}dz \
&= frac{1}{2+i}ze^{(2+i)z} - frac{1}{(2+i)^2}e^{(2+i)z} + C \
&= frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z} + C
end{align}
Taking the imaginary part of the above, we obtain
$$ operatorname{Re}left{ frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z}right} = frac25 ze^{2z}sin z - frac15 ze^{2z}cos z - frac{3}{25}e^{2z}sin z + frac{4}{25}e^{2z}cos z $$
answered Dec 25 '18 at 13:29
DylanDylan
13.7k31027
$endgroup$
add a comment |
$begingroup$
Note that
$$ int ze^{2z}sin z dz = operatorname{Im}left{int ze^{(2+i)z} dz right} $$
Using integration by parts
begin{align}
int z e^{(2+i)z} dz &= frac{1}{2+i}ze^{(2+i)z}- frac{1}{2+i}int e^{(2+i)z}dz \
&= frac{1}{2+i}ze^{(2+i)z} - frac{1}{(2+i)^2}e^{(2+i)z} + C \
&= frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z} + C
end{align}
Taking the imaginary part of the above, we obtain
$$ operatorname{Re}left{ frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z}right} = frac25 ze^{2z}sin z - frac15 ze^{2z}cos z - frac{3}{25}e^{2z}sin z + frac{4}{25}e^{2z}cos z $$
answered Dec 25 '18 at 13:29
DylanDylan
13.7k31027
$endgroup$
add a comment |
$begingroup$
Note that
$$ int ze^{2z}sin z dz = operatorname{Im}left{int ze^{(2+i)z} dz right} $$
Using integration by parts
begin{align}
int z e^{(2+i)z} dz &= frac{1}{2+i}ze^{(2+i)z}- frac{1}{2+i}int e^{(2+i)z}dz \
&= frac{1}{2+i}ze^{(2+i)z} - frac{1}{(2+i)^2}e^{(2+i)z} + C \
&= frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z} + C
end{align}
Taking the imaginary part of the above, we obtain
$$ operatorname{Re}left{ frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z}right} = frac25 ze^{2z}sin z - frac15 ze^{2z}cos z - frac{3}{25}e^{2z}sin z + frac{4}{25}e^{2z}cos z $$
answered Dec 25 '18 at 13:29
DylanDylan
13.7k31027
$endgroup$
Note that
$$ int ze^{2z}sin z dz = operatorname{Im}left{int ze^{(2+i)z} dz right} $$
Using integration by parts
begin{align}
int z e^{(2+i)z} dz &= frac{1}{2+i}ze^{(2+i)z}- frac{1}{2+i}int e^{(2+i)z}dz \
&= frac{1}{2+i}ze^{(2+i)z} - frac{1}{(2+i)^2}e^{(2+i)z} + C \
&= frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z} + C
end{align}
Taking the imaginary part of the above, we obtain
$$ operatorname{Re}left{ frac{2-i}{5}ze^{(2+i)z} - frac{3-4i}{25}e^{(2+i)z}right} = frac25 ze^{2z}sin z - frac15 ze^{2z}cos z - frac{3}{25}e^{2z}sin z + frac{4}{25}e^{2z}cos z $$
answered Dec 25 '18 at 13:29
DylanDylan
13.7k31027
answered Dec 25 '18 at 13:29
DylanDylan
13.7k31027
answered Dec 25 '18 at 13:29
DylanDylan
13.7k31027
answered Dec 25 '18 at 13:29
DylanDylan
13.7k31027
13.7k31027
add a comment |
add a comment |
$begingroup$
Use integration by parts two times
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 11:39
3
$begingroup$
Well what have you tried? Please show us your attempts, in MathJax, even if they are incorrect or don't lead anywhere. Perhaps you should start with integrating $zsin z$ first.
$endgroup$
– TheSimpliFire
Dec 25 '18 at 11:39