Expressing $int_0^{pi/2}frac{sin 2x}{x+1},dx$ using $A = int_0^pifrac{cos x}{(x+2)^2},dx$ [closed]
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If $$A = int_0^pifrac{cos x}{(x+2)^2},dx,$$
then find the value of $$int_0^{pi/2}frac{sin 2x}{x+1},dx$$ in terms of $A$.
calculus integration definite-integrals
edited Dec 13 '18 at 15:02
Martin Sleziak
44.8k9118272
asked Dec 13 '18 at 11:12
NoFluxGivenNoFluxGiven
113
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closed as off-topic by user10354138, amWhy, onurcanbektas, jameselmore, RRL Dec 13 '18 at 19:01
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." – user10354138, amWhy, onurcanbektas, jameselmore, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If $$A = int_0^pifrac{cos x}{(x+2)^2},dx,$$
then find the value of $$int_0^{pi/2}frac{sin 2x}{x+1},dx$$ in terms of $A$.
calculus integration definite-integrals
edited Dec 13 '18 at 15:02
Martin Sleziak
44.8k9118272
asked Dec 13 '18 at 11:12
NoFluxGivenNoFluxGiven
113
$endgroup$
closed as off-topic by user10354138, amWhy, onurcanbektas, jameselmore, RRL Dec 13 '18 at 19:01
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." – user10354138, amWhy, onurcanbektas, jameselmore, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
5
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Hint: Try an integration by parts for the first integral and try $u = 2x$ in your second integral
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– TheD0ubleT
Dec 13 '18 at 11:15
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Yeah! That works! Thank you so much. :))
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– NoFluxGiven
Dec 13 '18 at 11:46
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@NoFluxGiven If using the hint in the previous comment you were able to find solution, one possibility what to do would be to post it as an answer yourself. (Especially if you consider that the question might be useful for other users too. And, as a bonus, people will look at your solution and check whether it's ok.)
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– Martin Sleziak
Dec 13 '18 at 15:09
add a comment |
$begingroup$
If $$A = int_0^pifrac{cos x}{(x+2)^2},dx,$$
then find the value of $$int_0^{pi/2}frac{sin 2x}{x+1},dx$$ in terms of $A$.
calculus integration definite-integrals
edited Dec 13 '18 at 15:02
Martin Sleziak
44.8k9118272
asked Dec 13 '18 at 11:12
NoFluxGivenNoFluxGiven
113
$endgroup$
If $$A = int_0^pifrac{cos x}{(x+2)^2},dx,$$
then find the value of $$int_0^{pi/2}frac{sin 2x}{x+1},dx$$ in terms of $A$.
calculus integration definite-integrals
calculus integration definite-integrals
edited Dec 13 '18 at 15:02
Martin Sleziak
44.8k9118272
asked Dec 13 '18 at 11:12
NoFluxGivenNoFluxGiven
113
edited Dec 13 '18 at 15:02
Martin Sleziak
44.8k9118272
asked Dec 13 '18 at 11:12
NoFluxGivenNoFluxGiven
113
edited Dec 13 '18 at 15:02
Martin Sleziak
44.8k9118272
edited Dec 13 '18 at 15:02
Martin Sleziak
44.8k9118272
edited Dec 13 '18 at 15:02
Martin Sleziak
44.8k9118272
44.8k9118272
asked Dec 13 '18 at 11:12
NoFluxGivenNoFluxGiven
113
asked Dec 13 '18 at 11:12
NoFluxGivenNoFluxGiven
113
asked Dec 13 '18 at 11:12
NoFluxGivenNoFluxGiven
113
113
closed as off-topic by user10354138, amWhy, onurcanbektas, jameselmore, RRL Dec 13 '18 at 19:01
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." – user10354138, amWhy, onurcanbektas, jameselmore, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user10354138, amWhy, onurcanbektas, jameselmore, RRL Dec 13 '18 at 19:01
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." – user10354138, amWhy, onurcanbektas, jameselmore, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
Hint: Try an integration by parts for the first integral and try $u = 2x$ in your second integral
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– TheD0ubleT
Dec 13 '18 at 11:15
$begingroup$
Yeah! That works! Thank you so much. :))
$endgroup$
– NoFluxGiven
Dec 13 '18 at 11:46
$begingroup$
@NoFluxGiven If using the hint in the previous comment you were able to find solution, one possibility what to do would be to post it as an answer yourself. (Especially if you consider that the question might be useful for other users too. And, as a bonus, people will look at your solution and check whether it's ok.)
$endgroup$
– Martin Sleziak
Dec 13 '18 at 15:09
add a comment |
5
$begingroup$
Hint: Try an integration by parts for the first integral and try $u = 2x$ in your second integral
$endgroup$
– TheD0ubleT
Dec 13 '18 at 11:15
$begingroup$
Yeah! That works! Thank you so much. :))
$endgroup$
– NoFluxGiven
Dec 13 '18 at 11:46
$begingroup$
@NoFluxGiven If using the hint in the previous comment you were able to find solution, one possibility what to do would be to post it as an answer yourself. (Especially if you consider that the question might be useful for other users too. And, as a bonus, people will look at your solution and check whether it's ok.)
$endgroup$
– Martin Sleziak
Dec 13 '18 at 15:09
5
5
$begingroup$
Hint: Try an integration by parts for the first integral and try $u = 2x$ in your second integral
$endgroup$
– TheD0ubleT
Dec 13 '18 at 11:15
$begingroup$
Hint: Try an integration by parts for the first integral and try $u = 2x$ in your second integral
$endgroup$
– TheD0ubleT
Dec 13 '18 at 11:15
$begingroup$
Yeah! That works! Thank you so much. :))
$endgroup$
– NoFluxGiven
Dec 13 '18 at 11:46
$begingroup$
Yeah! That works! Thank you so much. :))
$endgroup$
– NoFluxGiven
Dec 13 '18 at 11:46
$begingroup$
@NoFluxGiven If using the hint in the previous comment you were able to find solution, one possibility what to do would be to post it as an answer yourself. (Especially if you consider that the question might be useful for other users too. And, as a bonus, people will look at your solution and check whether it's ok.)
$endgroup$
– Martin Sleziak
Dec 13 '18 at 15:09
$begingroup$
@NoFluxGiven If using the hint in the previous comment you were able to find solution, one possibility what to do would be to post it as an answer yourself. (Especially if you consider that the question might be useful for other users too. And, as a bonus, people will look at your solution and check whether it's ok.)
$endgroup$
– Martin Sleziak
Dec 13 '18 at 15:09
add a comment |
1 Answer
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As pointed out in the comments, the substitution $x mapsto 2x $ gives
$$ int_0^{pi/2} frac{sin 2x}{2x+2}2dx = int_0^pi frac{sin x}{x+2}dx $$
Integration by parts gives
$$ int_0^pi frac{sin x}{x+2}dx = -frac{cos x}{x+2}Biggvert_0^pi - int_0^pi frac{cos x}{(x+2)^2}dx = frac{1}{pi+2} + frac12 - A $$
answered Dec 13 '18 at 18:08
DylanDylan
12.8k31027
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As pointed out in the comments, the substitution $x mapsto 2x $ gives
$$ int_0^{pi/2} frac{sin 2x}{2x+2}2dx = int_0^pi frac{sin x}{x+2}dx $$
Integration by parts gives
$$ int_0^pi frac{sin x}{x+2}dx = -frac{cos x}{x+2}Biggvert_0^pi - int_0^pi frac{cos x}{(x+2)^2}dx = frac{1}{pi+2} + frac12 - A $$
answered Dec 13 '18 at 18:08
DylanDylan
12.8k31027
$endgroup$
add a comment |
$begingroup$
As pointed out in the comments, the substitution $x mapsto 2x $ gives
$$ int_0^{pi/2} frac{sin 2x}{2x+2}2dx = int_0^pi frac{sin x}{x+2}dx $$
Integration by parts gives
$$ int_0^pi frac{sin x}{x+2}dx = -frac{cos x}{x+2}Biggvert_0^pi - int_0^pi frac{cos x}{(x+2)^2}dx = frac{1}{pi+2} + frac12 - A $$
answered Dec 13 '18 at 18:08
DylanDylan
12.8k31027
$endgroup$
add a comment |
$begingroup$
As pointed out in the comments, the substitution $x mapsto 2x $ gives
$$ int_0^{pi/2} frac{sin 2x}{2x+2}2dx = int_0^pi frac{sin x}{x+2}dx $$
Integration by parts gives
$$ int_0^pi frac{sin x}{x+2}dx = -frac{cos x}{x+2}Biggvert_0^pi - int_0^pi frac{cos x}{(x+2)^2}dx = frac{1}{pi+2} + frac12 - A $$
answered Dec 13 '18 at 18:08
DylanDylan
12.8k31027
$endgroup$
As pointed out in the comments, the substitution $x mapsto 2x $ gives
$$ int_0^{pi/2} frac{sin 2x}{2x+2}2dx = int_0^pi frac{sin x}{x+2}dx $$
Integration by parts gives
$$ int_0^pi frac{sin x}{x+2}dx = -frac{cos x}{x+2}Biggvert_0^pi - int_0^pi frac{cos x}{(x+2)^2}dx = frac{1}{pi+2} + frac12 - A $$
answered Dec 13 '18 at 18:08
DylanDylan
12.8k31027
answered Dec 13 '18 at 18:08
DylanDylan
12.8k31027
answered Dec 13 '18 at 18:08
DylanDylan
12.8k31027
answered Dec 13 '18 at 18:08
DylanDylan
12.8k31027
12.8k31027
add a comment |
add a comment |
5
$begingroup$
Hint: Try an integration by parts for the first integral and try $u = 2x$ in your second integral
$endgroup$
– TheD0ubleT
Dec 13 '18 at 11:15
$begingroup$
Yeah! That works! Thank you so much. :))
$endgroup$
– NoFluxGiven
Dec 13 '18 at 11:46
$begingroup$
@NoFluxGiven If using the hint in the previous comment you were able to find solution, one possibility what to do would be to post it as an answer yourself. (Especially if you consider that the question might be useful for other users too. And, as a bonus, people will look at your solution and check whether it's ok.)
$endgroup$
– Martin Sleziak
Dec 13 '18 at 15:09