Solve $int e^x sin(9x),dx$ using integration by parts
I have this Integration by Parts question that I can't seem to find an answer to.
The question is:
$$int e^x
sin(9x),dx$$
I used u-substitution:
$$u=e^x,du=e^x,dx$$
$$dv=sin(9x),dx, v=-frac{1}{9}cos(9x)$$
Then I got:
$$-frac{1}{9}e^xcos(9x)+frac{1}{9}int e^xcos(9x),dx$$
After a second integration I used:
$$u=e^x, du=e^x,dx$$
$$dv=cos(9x),dx, v=frac{1}{9}sin(9x)$$
Furthermore,
$$int e^xsin(9x)dx=-frac{1}{9} e^xcos(9x)+frac{1}{9}left(frac{1}{9} e^xsin(9x)-frac{1}{9}int e^xsin(9x),dxright)$$
$$int e^xsin(9x)dx=-frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)-frac{1}{81}int e^xsin(9x),dx$$
I'm stuck and I'm not sure exactly what to do after this.
Any help would be very grateful..thanks!
calculus integration indefinite-integrals
edited Nov 25 at 8:52
Martin Sleziak
44.6k7115270
asked Mar 8 '14 at 0:59
AlecLeonK
328212
add a comment |
I have this Integration by Parts question that I can't seem to find an answer to.
The question is:
$$int e^x
sin(9x),dx$$
I used u-substitution:
$$u=e^x,du=e^x,dx$$
$$dv=sin(9x),dx, v=-frac{1}{9}cos(9x)$$
Then I got:
$$-frac{1}{9}e^xcos(9x)+frac{1}{9}int e^xcos(9x),dx$$
After a second integration I used:
$$u=e^x, du=e^x,dx$$
$$dv=cos(9x),dx, v=frac{1}{9}sin(9x)$$
Furthermore,
$$int e^xsin(9x)dx=-frac{1}{9} e^xcos(9x)+frac{1}{9}left(frac{1}{9} e^xsin(9x)-frac{1}{9}int e^xsin(9x),dxright)$$
$$int e^xsin(9x)dx=-frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)-frac{1}{81}int e^xsin(9x),dx$$
I'm stuck and I'm not sure exactly what to do after this.
Any help would be very grateful..thanks!
calculus integration indefinite-integrals
edited Nov 25 at 8:52
Martin Sleziak
44.6k7115270
asked Mar 8 '14 at 0:59
AlecLeonK
328212
1
For clear expression of any trigonometric equations, be sure to include before cosine and sine.
– NasuSama
Mar 8 '14 at 1:04
add a comment |
I have this Integration by Parts question that I can't seem to find an answer to.
The question is:
$$int e^x
sin(9x),dx$$
I used u-substitution:
$$u=e^x,du=e^x,dx$$
$$dv=sin(9x),dx, v=-frac{1}{9}cos(9x)$$
Then I got:
$$-frac{1}{9}e^xcos(9x)+frac{1}{9}int e^xcos(9x),dx$$
After a second integration I used:
$$u=e^x, du=e^x,dx$$
$$dv=cos(9x),dx, v=frac{1}{9}sin(9x)$$
Furthermore,
$$int e^xsin(9x)dx=-frac{1}{9} e^xcos(9x)+frac{1}{9}left(frac{1}{9} e^xsin(9x)-frac{1}{9}int e^xsin(9x),dxright)$$
$$int e^xsin(9x)dx=-frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)-frac{1}{81}int e^xsin(9x),dx$$
I'm stuck and I'm not sure exactly what to do after this.
Any help would be very grateful..thanks!
calculus integration indefinite-integrals
edited Nov 25 at 8:52
Martin Sleziak
44.6k7115270
asked Mar 8 '14 at 0:59
AlecLeonK
328212
I have this Integration by Parts question that I can't seem to find an answer to.
The question is:
$$int e^x
sin(9x),dx$$
I used u-substitution:
$$u=e^x,du=e^x,dx$$
$$dv=sin(9x),dx, v=-frac{1}{9}cos(9x)$$
Then I got:
$$-frac{1}{9}e^xcos(9x)+frac{1}{9}int e^xcos(9x),dx$$
After a second integration I used:
$$u=e^x, du=e^x,dx$$
$$dv=cos(9x),dx, v=frac{1}{9}sin(9x)$$
Furthermore,
$$int e^xsin(9x)dx=-frac{1}{9} e^xcos(9x)+frac{1}{9}left(frac{1}{9} e^xsin(9x)-frac{1}{9}int e^xsin(9x),dxright)$$
$$int e^xsin(9x)dx=-frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)-frac{1}{81}int e^xsin(9x),dx$$
I'm stuck and I'm not sure exactly what to do after this.
Any help would be very grateful..thanks!
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Nov 25 at 8:52
Martin Sleziak
44.6k7115270
asked Mar 8 '14 at 0:59
AlecLeonK
328212
edited Nov 25 at 8:52
Martin Sleziak
44.6k7115270
asked Mar 8 '14 at 0:59
AlecLeonK
328212
edited Nov 25 at 8:52
Martin Sleziak
44.6k7115270
edited Nov 25 at 8:52
Martin Sleziak
44.6k7115270
edited Nov 25 at 8:52
Martin Sleziak
44.6k7115270
44.6k7115270
asked Mar 8 '14 at 0:59
AlecLeonK
328212
asked Mar 8 '14 at 0:59
AlecLeonK
328212
asked Mar 8 '14 at 0:59
AlecLeonK
328212
328212
1
For clear expression of any trigonometric equations, be sure to include before cosine and sine.
– NasuSama
Mar 8 '14 at 1:04
add a comment |
1
For clear expression of any trigonometric equations, be sure to include before cosine and sine.
– NasuSama
Mar 8 '14 at 1:04
1
1
For clear expression of any trigonometric equations, be sure to include before cosine and sine.
– NasuSama
Mar 8 '14 at 1:04
For clear expression of any trigonometric equations, be sure to include before cosine and sine.
– NasuSama
Mar 8 '14 at 1:04
add a comment |
4 Answers
4
active
oldest
votes
More simply and without integration by parts we have
$$int e^xsin(9x)dx=operatorname{Im}int e^{(1+9i)x}dx=operatorname{Im}left(frac1{1+9i}e^{(1+9i)x}right)=operatorname{Im}left(frac1{82}(1-9i)e^{(1+9i)x}right)$$
Now develop and take the imaginary part.
edited Mar 8 '14 at 1:33
NasuSama
2,5101035
Where did you learn this? I've never seen it answered like that before
– AlecLeonK
Mar 8 '14 at 1:22
1
@AlecLeonK First, write $e^{(1 + 9i)x}$ as $e^xe^{9ix}$. Then, for $e^{9ix}$, use Euler's identity and find the imaginary part of that expression. For more information about this method, see here.
– NasuSama
Mar 8 '14 at 1:31
add a comment |
From where you left off, let
$$s = int e^xsin(9x),dx$$
Then, we have
$$s = -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) - dfrac{1}{81}s$$
Solving for $s$ gives
$$begin{aligned}
dfrac{82}{81}s &= -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x)\
s &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)\
int e^xsin(9x),dx &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)
end{aligned}$$
answered Mar 8 '14 at 1:03
NasuSama
2,5101035
I typed it in and I see the logic behind it but for some reason it says it's wrong. I understand what you did though.
– AlecLeonK
Mar 8 '14 at 1:11
I'm very glad to hear that. I think you ask this question that comes from the online homework, am I right?
– NasuSama
Mar 8 '14 at 1:22
Yes it does, do you know why it might be wrong?
– AlecLeonK
Mar 8 '14 at 1:40
I don't know. Assuming your online HW system does not give away answer after exhausted amount of attempts or after HW deadline, I would assume that the HW system computed the integral in the complicated way without using the elementary method you learned (I guessed).
– NasuSama
Mar 8 '14 at 1:42
Would it make a difference if I make the orginal 'u'=sin(9x) instead of e^x?
– AlecLeonK
Mar 8 '14 at 1:46
add a comment |
You have
$$int e^xsin(9x)dx=−frac{1}{9}e^xcos(9x)+frac{1}{81} e^xsin(9x)−frac{1}{81}int e^xsin(9x)dx$$ Then bring $−frac{1}{81}int e^xsin(9x)dx$ to the other side and you get
$$frac{82}{81}int e^xsin(9x)dx=−frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)$$
edited Mar 8 '14 at 1:07
AlecLeonK
328212
answered Mar 8 '14 at 1:03
Kaladin
1,230814
add a comment |
We can take a more general approach, and find
$$I=int e^{ax}sin bx mathrm{d}x$$
Integration by parts:
$$mathrm{d}v=sin bx mathrm{d}xRightarrow v=-frac1bcos bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug it in:
$$I=-frac{e^{ax}}bcos bx+frac abint e^{ax}cos bx mathrm{d}x$$
Integration by parts round $2$:
$$mathrm{d}v=cos bx mathrm{d}xRightarrow v=frac1bsin bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug in:
$$I=-frac{e^{ax}}bcos bx+frac abbigg(frac{e^{ax}}{b}sin bx-frac abint e^{ax}sin bx mathrm{d}xbigg)$$
Take note of the $int e^{ax}sin bx mathrm{d}x$ term:
$$I=-frac{e^{ax}}bcos bx+frac{ae^{ax}}{b^2}sin bx-frac{a^2}{b^2}I$$
$$bigg(1+frac{a^2}{b^2}bigg)I=frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{b^2}{a^2+b^2}frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{e^{ax}}{a^2+b^2}big(asin bx-bcos bxbig)+C$$
Isn't that cool? This is called a reduction formula, and there are many more of them.
answered Nov 24 at 19:41
clathratus
2,842327
1
Perhaps it's worth mentioning that there are several posts about this type of integrals on this site. Such as Integrate $e^{ax}sin(bx)?$ (and other posts linked there).This post might also be of interest: Integration by parts: $int e^{ax}cos(bx),dx$ (together with the linked questions.)
– Martin Sleziak
Nov 25 at 8:56
add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
More simply and without integration by parts we have
$$int e^xsin(9x)dx=operatorname{Im}int e^{(1+9i)x}dx=operatorname{Im}left(frac1{1+9i}e^{(1+9i)x}right)=operatorname{Im}left(frac1{82}(1-9i)e^{(1+9i)x}right)$$
Now develop and take the imaginary part.
edited Mar 8 '14 at 1:33
NasuSama
2,5101035
Where did you learn this? I've never seen it answered like that before
– AlecLeonK
Mar 8 '14 at 1:22
1
@AlecLeonK First, write $e^{(1 + 9i)x}$ as $e^xe^{9ix}$. Then, for $e^{9ix}$, use Euler's identity and find the imaginary part of that expression. For more information about this method, see here.
– NasuSama
Mar 8 '14 at 1:31
add a comment |
More simply and without integration by parts we have
$$int e^xsin(9x)dx=operatorname{Im}int e^{(1+9i)x}dx=operatorname{Im}left(frac1{1+9i}e^{(1+9i)x}right)=operatorname{Im}left(frac1{82}(1-9i)e^{(1+9i)x}right)$$
Now develop and take the imaginary part.
edited Mar 8 '14 at 1:33
NasuSama
2,5101035
Where did you learn this? I've never seen it answered like that before
– AlecLeonK
Mar 8 '14 at 1:22
1
@AlecLeonK First, write $e^{(1 + 9i)x}$ as $e^xe^{9ix}$. Then, for $e^{9ix}$, use Euler's identity and find the imaginary part of that expression. For more information about this method, see here.
– NasuSama
Mar 8 '14 at 1:31
add a comment |
More simply and without integration by parts we have
$$int e^xsin(9x)dx=operatorname{Im}int e^{(1+9i)x}dx=operatorname{Im}left(frac1{1+9i}e^{(1+9i)x}right)=operatorname{Im}left(frac1{82}(1-9i)e^{(1+9i)x}right)$$
Now develop and take the imaginary part.
edited Mar 8 '14 at 1:33
NasuSama
2,5101035
More simply and without integration by parts we have
$$int e^xsin(9x)dx=operatorname{Im}int e^{(1+9i)x}dx=operatorname{Im}left(frac1{1+9i}e^{(1+9i)x}right)=operatorname{Im}left(frac1{82}(1-9i)e^{(1+9i)x}right)$$
Now develop and take the imaginary part.
edited Mar 8 '14 at 1:33
NasuSama
2,5101035
edited Mar 8 '14 at 1:33
NasuSama
2,5101035
edited Mar 8 '14 at 1:33
NasuSama
2,5101035
edited Mar 8 '14 at 1:33
NasuSama
2,5101035
2,5101035
answered Mar 8 '14 at 1:19
user63181
Where did you learn this? I've never seen it answered like that before
– AlecLeonK
Mar 8 '14 at 1:22
1
@AlecLeonK First, write $e^{(1 + 9i)x}$ as $e^xe^{9ix}$. Then, for $e^{9ix}$, use Euler's identity and find the imaginary part of that expression. For more information about this method, see here.
– NasuSama
Mar 8 '14 at 1:31
add a comment |
Where did you learn this? I've never seen it answered like that before
– AlecLeonK
Mar 8 '14 at 1:22
1
@AlecLeonK First, write $e^{(1 + 9i)x}$ as $e^xe^{9ix}$. Then, for $e^{9ix}$, use Euler's identity and find the imaginary part of that expression. For more information about this method, see here.
– NasuSama
Mar 8 '14 at 1:31
Where did you learn this? I've never seen it answered like that before
– AlecLeonK
Mar 8 '14 at 1:22
Where did you learn this? I've never seen it answered like that before
– AlecLeonK
Mar 8 '14 at 1:22
1
1
@AlecLeonK First, write $e^{(1 + 9i)x}$ as $e^xe^{9ix}$. Then, for $e^{9ix}$, use Euler's identity and find the imaginary part of that expression. For more information about this method, see here.
– NasuSama
Mar 8 '14 at 1:31
@AlecLeonK First, write $e^{(1 + 9i)x}$ as $e^xe^{9ix}$. Then, for $e^{9ix}$, use Euler's identity and find the imaginary part of that expression. For more information about this method, see here.
– NasuSama
Mar 8 '14 at 1:31
add a comment |
From where you left off, let
$$s = int e^xsin(9x),dx$$
Then, we have
$$s = -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) - dfrac{1}{81}s$$
Solving for $s$ gives
$$begin{aligned}
dfrac{82}{81}s &= -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x)\
s &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)\
int e^xsin(9x),dx &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)
end{aligned}$$
answered Mar 8 '14 at 1:03
NasuSama
2,5101035
I typed it in and I see the logic behind it but for some reason it says it's wrong. I understand what you did though.
– AlecLeonK
Mar 8 '14 at 1:11
I'm very glad to hear that. I think you ask this question that comes from the online homework, am I right?
– NasuSama
Mar 8 '14 at 1:22
Yes it does, do you know why it might be wrong?
– AlecLeonK
Mar 8 '14 at 1:40
I don't know. Assuming your online HW system does not give away answer after exhausted amount of attempts or after HW deadline, I would assume that the HW system computed the integral in the complicated way without using the elementary method you learned (I guessed).
– NasuSama
Mar 8 '14 at 1:42
Would it make a difference if I make the orginal 'u'=sin(9x) instead of e^x?
– AlecLeonK
Mar 8 '14 at 1:46
add a comment |
From where you left off, let
$$s = int e^xsin(9x),dx$$
Then, we have
$$s = -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) - dfrac{1}{81}s$$
Solving for $s$ gives
$$begin{aligned}
dfrac{82}{81}s &= -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x)\
s &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)\
int e^xsin(9x),dx &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)
end{aligned}$$
answered Mar 8 '14 at 1:03
NasuSama
2,5101035
I typed it in and I see the logic behind it but for some reason it says it's wrong. I understand what you did though.
– AlecLeonK
Mar 8 '14 at 1:11
I'm very glad to hear that. I think you ask this question that comes from the online homework, am I right?
– NasuSama
Mar 8 '14 at 1:22
Yes it does, do you know why it might be wrong?
– AlecLeonK
Mar 8 '14 at 1:40
I don't know. Assuming your online HW system does not give away answer after exhausted amount of attempts or after HW deadline, I would assume that the HW system computed the integral in the complicated way without using the elementary method you learned (I guessed).
– NasuSama
Mar 8 '14 at 1:42
Would it make a difference if I make the orginal 'u'=sin(9x) instead of e^x?
– AlecLeonK
Mar 8 '14 at 1:46
add a comment |
From where you left off, let
$$s = int e^xsin(9x),dx$$
Then, we have
$$s = -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) - dfrac{1}{81}s$$
Solving for $s$ gives
$$begin{aligned}
dfrac{82}{81}s &= -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x)\
s &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)\
int e^xsin(9x),dx &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)
end{aligned}$$
answered Mar 8 '14 at 1:03
NasuSama
2,5101035
From where you left off, let
$$s = int e^xsin(9x),dx$$
Then, we have
$$s = -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) - dfrac{1}{81}s$$
Solving for $s$ gives
$$begin{aligned}
dfrac{82}{81}s &= -dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x)\
s &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)\
int e^xsin(9x),dx &= dfrac{81}{82}left(-dfrac{1}{9}e^xcos(9x) + dfrac{1}{81}e^xsin(9x) right)
end{aligned}$$
answered Mar 8 '14 at 1:03
NasuSama
2,5101035
answered Mar 8 '14 at 1:03
NasuSama
2,5101035
answered Mar 8 '14 at 1:03
NasuSama
2,5101035
answered Mar 8 '14 at 1:03
NasuSama
2,5101035
2,5101035
I typed it in and I see the logic behind it but for some reason it says it's wrong. I understand what you did though.
– AlecLeonK
Mar 8 '14 at 1:11
I'm very glad to hear that. I think you ask this question that comes from the online homework, am I right?
– NasuSama
Mar 8 '14 at 1:22
Yes it does, do you know why it might be wrong?
– AlecLeonK
Mar 8 '14 at 1:40
I don't know. Assuming your online HW system does not give away answer after exhausted amount of attempts or after HW deadline, I would assume that the HW system computed the integral in the complicated way without using the elementary method you learned (I guessed).
– NasuSama
Mar 8 '14 at 1:42
Would it make a difference if I make the orginal 'u'=sin(9x) instead of e^x?
– AlecLeonK
Mar 8 '14 at 1:46
add a comment |
I typed it in and I see the logic behind it but for some reason it says it's wrong. I understand what you did though.
– AlecLeonK
Mar 8 '14 at 1:11
I'm very glad to hear that. I think you ask this question that comes from the online homework, am I right?
– NasuSama
Mar 8 '14 at 1:22
Yes it does, do you know why it might be wrong?
– AlecLeonK
Mar 8 '14 at 1:40
I don't know. Assuming your online HW system does not give away answer after exhausted amount of attempts or after HW deadline, I would assume that the HW system computed the integral in the complicated way without using the elementary method you learned (I guessed).
– NasuSama
Mar 8 '14 at 1:42
Would it make a difference if I make the orginal 'u'=sin(9x) instead of e^x?
– AlecLeonK
Mar 8 '14 at 1:46
I typed it in and I see the logic behind it but for some reason it says it's wrong. I understand what you did though.
– AlecLeonK
Mar 8 '14 at 1:11
I typed it in and I see the logic behind it but for some reason it says it's wrong. I understand what you did though.
– AlecLeonK
Mar 8 '14 at 1:11
I'm very glad to hear that. I think you ask this question that comes from the online homework, am I right?
– NasuSama
Mar 8 '14 at 1:22
I'm very glad to hear that. I think you ask this question that comes from the online homework, am I right?
– NasuSama
Mar 8 '14 at 1:22
Yes it does, do you know why it might be wrong?
– AlecLeonK
Mar 8 '14 at 1:40
Yes it does, do you know why it might be wrong?
– AlecLeonK
Mar 8 '14 at 1:40
I don't know. Assuming your online HW system does not give away answer after exhausted amount of attempts or after HW deadline, I would assume that the HW system computed the integral in the complicated way without using the elementary method you learned (I guessed).
– NasuSama
Mar 8 '14 at 1:42
I don't know. Assuming your online HW system does not give away answer after exhausted amount of attempts or after HW deadline, I would assume that the HW system computed the integral in the complicated way without using the elementary method you learned (I guessed).
– NasuSama
Mar 8 '14 at 1:42
Would it make a difference if I make the orginal 'u'=sin(9x) instead of e^x?
– AlecLeonK
Mar 8 '14 at 1:46
Would it make a difference if I make the orginal 'u'=sin(9x) instead of e^x?
– AlecLeonK
Mar 8 '14 at 1:46
add a comment |
You have
$$int e^xsin(9x)dx=−frac{1}{9}e^xcos(9x)+frac{1}{81} e^xsin(9x)−frac{1}{81}int e^xsin(9x)dx$$ Then bring $−frac{1}{81}int e^xsin(9x)dx$ to the other side and you get
$$frac{82}{81}int e^xsin(9x)dx=−frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)$$
edited Mar 8 '14 at 1:07
AlecLeonK
328212
answered Mar 8 '14 at 1:03
Kaladin
1,230814
add a comment |
You have
$$int e^xsin(9x)dx=−frac{1}{9}e^xcos(9x)+frac{1}{81} e^xsin(9x)−frac{1}{81}int e^xsin(9x)dx$$ Then bring $−frac{1}{81}int e^xsin(9x)dx$ to the other side and you get
$$frac{82}{81}int e^xsin(9x)dx=−frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)$$
edited Mar 8 '14 at 1:07
AlecLeonK
328212
answered Mar 8 '14 at 1:03
Kaladin
1,230814
add a comment |
You have
$$int e^xsin(9x)dx=−frac{1}{9}e^xcos(9x)+frac{1}{81} e^xsin(9x)−frac{1}{81}int e^xsin(9x)dx$$ Then bring $−frac{1}{81}int e^xsin(9x)dx$ to the other side and you get
$$frac{82}{81}int e^xsin(9x)dx=−frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)$$
edited Mar 8 '14 at 1:07
AlecLeonK
328212
answered Mar 8 '14 at 1:03
Kaladin
1,230814
You have
$$int e^xsin(9x)dx=−frac{1}{9}e^xcos(9x)+frac{1}{81} e^xsin(9x)−frac{1}{81}int e^xsin(9x)dx$$ Then bring $−frac{1}{81}int e^xsin(9x)dx$ to the other side and you get
$$frac{82}{81}int e^xsin(9x)dx=−frac{1}{9} e^xcos(9x)+frac{1}{81} e^xsin(9x)$$
edited Mar 8 '14 at 1:07
AlecLeonK
328212
answered Mar 8 '14 at 1:03
Kaladin
1,230814
edited Mar 8 '14 at 1:07
AlecLeonK
328212
edited Mar 8 '14 at 1:07
AlecLeonK
328212
edited Mar 8 '14 at 1:07
AlecLeonK
328212
328212
answered Mar 8 '14 at 1:03
Kaladin
1,230814
answered Mar 8 '14 at 1:03
Kaladin
1,230814
answered Mar 8 '14 at 1:03
Kaladin
1,230814
1,230814
add a comment |
add a comment |
We can take a more general approach, and find
$$I=int e^{ax}sin bx mathrm{d}x$$
Integration by parts:
$$mathrm{d}v=sin bx mathrm{d}xRightarrow v=-frac1bcos bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug it in:
$$I=-frac{e^{ax}}bcos bx+frac abint e^{ax}cos bx mathrm{d}x$$
Integration by parts round $2$:
$$mathrm{d}v=cos bx mathrm{d}xRightarrow v=frac1bsin bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug in:
$$I=-frac{e^{ax}}bcos bx+frac abbigg(frac{e^{ax}}{b}sin bx-frac abint e^{ax}sin bx mathrm{d}xbigg)$$
Take note of the $int e^{ax}sin bx mathrm{d}x$ term:
$$I=-frac{e^{ax}}bcos bx+frac{ae^{ax}}{b^2}sin bx-frac{a^2}{b^2}I$$
$$bigg(1+frac{a^2}{b^2}bigg)I=frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{b^2}{a^2+b^2}frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{e^{ax}}{a^2+b^2}big(asin bx-bcos bxbig)+C$$
Isn't that cool? This is called a reduction formula, and there are many more of them.
answered Nov 24 at 19:41
clathratus
2,842327
1
Perhaps it's worth mentioning that there are several posts about this type of integrals on this site. Such as Integrate $e^{ax}sin(bx)?$ (and other posts linked there).This post might also be of interest: Integration by parts: $int e^{ax}cos(bx),dx$ (together with the linked questions.)
– Martin Sleziak
Nov 25 at 8:56
add a comment |
We can take a more general approach, and find
$$I=int e^{ax}sin bx mathrm{d}x$$
Integration by parts:
$$mathrm{d}v=sin bx mathrm{d}xRightarrow v=-frac1bcos bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug it in:
$$I=-frac{e^{ax}}bcos bx+frac abint e^{ax}cos bx mathrm{d}x$$
Integration by parts round $2$:
$$mathrm{d}v=cos bx mathrm{d}xRightarrow v=frac1bsin bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug in:
$$I=-frac{e^{ax}}bcos bx+frac abbigg(frac{e^{ax}}{b}sin bx-frac abint e^{ax}sin bx mathrm{d}xbigg)$$
Take note of the $int e^{ax}sin bx mathrm{d}x$ term:
$$I=-frac{e^{ax}}bcos bx+frac{ae^{ax}}{b^2}sin bx-frac{a^2}{b^2}I$$
$$bigg(1+frac{a^2}{b^2}bigg)I=frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{b^2}{a^2+b^2}frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{e^{ax}}{a^2+b^2}big(asin bx-bcos bxbig)+C$$
Isn't that cool? This is called a reduction formula, and there are many more of them.
answered Nov 24 at 19:41
clathratus
2,842327
1
Perhaps it's worth mentioning that there are several posts about this type of integrals on this site. Such as Integrate $e^{ax}sin(bx)?$ (and other posts linked there).This post might also be of interest: Integration by parts: $int e^{ax}cos(bx),dx$ (together with the linked questions.)
– Martin Sleziak
Nov 25 at 8:56
add a comment |
We can take a more general approach, and find
$$I=int e^{ax}sin bx mathrm{d}x$$
Integration by parts:
$$mathrm{d}v=sin bx mathrm{d}xRightarrow v=-frac1bcos bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug it in:
$$I=-frac{e^{ax}}bcos bx+frac abint e^{ax}cos bx mathrm{d}x$$
Integration by parts round $2$:
$$mathrm{d}v=cos bx mathrm{d}xRightarrow v=frac1bsin bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug in:
$$I=-frac{e^{ax}}bcos bx+frac abbigg(frac{e^{ax}}{b}sin bx-frac abint e^{ax}sin bx mathrm{d}xbigg)$$
Take note of the $int e^{ax}sin bx mathrm{d}x$ term:
$$I=-frac{e^{ax}}bcos bx+frac{ae^{ax}}{b^2}sin bx-frac{a^2}{b^2}I$$
$$bigg(1+frac{a^2}{b^2}bigg)I=frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{b^2}{a^2+b^2}frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{e^{ax}}{a^2+b^2}big(asin bx-bcos bxbig)+C$$
Isn't that cool? This is called a reduction formula, and there are many more of them.
answered Nov 24 at 19:41
clathratus
2,842327
We can take a more general approach, and find
$$I=int e^{ax}sin bx mathrm{d}x$$
Integration by parts:
$$mathrm{d}v=sin bx mathrm{d}xRightarrow v=-frac1bcos bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug it in:
$$I=-frac{e^{ax}}bcos bx+frac abint e^{ax}cos bx mathrm{d}x$$
Integration by parts round $2$:
$$mathrm{d}v=cos bx mathrm{d}xRightarrow v=frac1bsin bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
Plug in:
$$I=-frac{e^{ax}}bcos bx+frac abbigg(frac{e^{ax}}{b}sin bx-frac abint e^{ax}sin bx mathrm{d}xbigg)$$
Take note of the $int e^{ax}sin bx mathrm{d}x$ term:
$$I=-frac{e^{ax}}bcos bx+frac{ae^{ax}}{b^2}sin bx-frac{a^2}{b^2}I$$
$$bigg(1+frac{a^2}{b^2}bigg)I=frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{b^2}{a^2+b^2}frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{e^{ax}}{a^2+b^2}big(asin bx-bcos bxbig)+C$$
Isn't that cool? This is called a reduction formula, and there are many more of them.
answered Nov 24 at 19:41
clathratus
2,842327
answered Nov 24 at 19:41
clathratus
2,842327
answered Nov 24 at 19:41
clathratus
2,842327
answered Nov 24 at 19:41
clathratus
2,842327
2,842327
1
Perhaps it's worth mentioning that there are several posts about this type of integrals on this site. Such as Integrate $e^{ax}sin(bx)?$ (and other posts linked there).This post might also be of interest: Integration by parts: $int e^{ax}cos(bx),dx$ (together with the linked questions.)
– Martin Sleziak
Nov 25 at 8:56
add a comment |
1
Perhaps it's worth mentioning that there are several posts about this type of integrals on this site. Such as Integrate $e^{ax}sin(bx)?$ (and other posts linked there).This post might also be of interest: Integration by parts: $int e^{ax}cos(bx),dx$ (together with the linked questions.)
– Martin Sleziak
Nov 25 at 8:56
1
1
Perhaps it's worth mentioning that there are several posts about this type of integrals on this site. Such as Integrate $e^{ax}sin(bx)?$ (and other posts linked there).This post might also be of interest: Integration by parts: $int e^{ax}cos(bx),dx$ (together with the linked questions.)
– Martin Sleziak
Nov 25 at 8:56
Perhaps it's worth mentioning that there are several posts about this type of integrals on this site. Such as Integrate $e^{ax}sin(bx)?$ (and other posts linked there).This post might also be of interest: Integration by parts: $int e^{ax}cos(bx),dx$ (together with the linked questions.)
– Martin Sleziak
Nov 25 at 8:56
add a comment |
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For clear expression of any trigonometric equations, be sure to include before cosine and sine.
– NasuSama
Mar 8 '14 at 1:04