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Evaluating the integral $int frac1(2+3cos x)^2mathrm dx$
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up vote
3
down vote
favorite
Please someone give me an idea to evaluate
this: $$int frac1(2+3cos x)^2mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $cos^2x$ does not work, so help me here.
calculus integration indefinite-integrals
edited Aug 3 at 18:47
Abcd
2,3151524
asked Aug 3 at 7:39
Ritik
4310
add a comment |Â
up vote
3
down vote
favorite
Please someone give me an idea to evaluate
this: $$int frac1(2+3cos x)^2mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $cos^2x$ does not work, so help me here.
calculus integration indefinite-integrals
edited Aug 3 at 18:47
Abcd
2,3151524
asked Aug 3 at 7:39
Ritik
4310
in.answers.yahoo.com/question/…
– lab bhattacharjee
Aug 3 at 7:41
Hint Multiply the integrand by $fracsec^2 xsec^2 x$.
– Travis
Aug 3 at 8:26
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Please someone give me an idea to evaluate
this: $$int frac1(2+3cos x)^2mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $cos^2x$ does not work, so help me here.
calculus integration indefinite-integrals
edited Aug 3 at 18:47
Abcd
2,3151524
asked Aug 3 at 7:39
Ritik
4310
Please someone give me an idea to evaluate
this: $$int frac1(2+3cos x)^2mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $cos^2x$ does not work, so help me here.
calculus integration indefinite-integrals
edited Aug 3 at 18:47
Abcd
2,3151524
asked Aug 3 at 7:39
Ritik
4310
edited Aug 3 at 18:47
Abcd
2,3151524
edited Aug 3 at 18:47
Abcd
2,3151524
edited Aug 3 at 18:47
Abcd
2,3151524
2,3151524
asked Aug 3 at 7:39
Ritik
4310
asked Aug 3 at 7:39
Ritik
4310
asked Aug 3 at 7:39
Ritik
4310
4310
in.answers.yahoo.com/question/…
– lab bhattacharjee
Aug 3 at 7:41
Hint Multiply the integrand by $fracsec^2 xsec^2 x$.
– Travis
Aug 3 at 8:26
add a comment |Â
in.answers.yahoo.com/question/…
– lab bhattacharjee
Aug 3 at 7:41
Hint Multiply the integrand by $fracsec^2 xsec^2 x$.
– Travis
Aug 3 at 8:26
in.answers.yahoo.com/question/…
– lab bhattacharjee
Aug 3 at 7:41
in.answers.yahoo.com/question/…
– lab bhattacharjee
Aug 3 at 7:41
Hint Multiply the integrand by $fracsec^2 xsec^2 x$.
– Travis
Aug 3 at 8:26
Hint Multiply the integrand by $fracsec^2 xsec^2 x$.
– Travis
Aug 3 at 8:26
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
Integrating by parts,
$$intdfracdx(a+bcos x)^2=intdfracsin x(a+bcos x)^2cdotdfrac1sin xdx$$
$$=dfrac1sin xintdfracsin x(a+bcos x)^2dx-left(dfracd(1/sin x)dxcdotintdfracsin x(a+bcos x)^2dxright)dx$$
$$=dfrac1bsin x(a+bcos x)+intdfraccos xb(1-cos^2x)(a+bcos x),dx$$
Use Partial Fraction Decomposition,
$$dfraccos x(1-cos^2x)(a+bcos x)=dfrac A1+cos x+dfrac B1-cos x+dfrac Ca+bcos x$$
The first two integral can be managed easily, for the last use Weierstrass substitution
edited Aug 3 at 8:46
Batominovski
22.6k22776
answered Aug 3 at 8:26
lab bhattacharjee
214k14152263
add a comment |Â
up vote
2
down vote
Thanks guys for answering my question .
I researched on it and found out a way to solve which I liked much.
So I want share to you guys too
Please have a look .
1
The idea is beautiful, you just reversed $2+3cos x$ with $3+2cos x$ by mistake. Good job!
– Zacky
Aug 3 at 12:37
Can you please elaborate where I did it
– Ritik
Aug 3 at 12:42
1
third row, after you differentiated $A$. Nevermind, its correct. Im tired.
– Zacky
Aug 3 at 12:44
1
Sir or mam when you will differentiate that A then You will get there 2 cosx +3 that was intentional
– Ritik
Aug 3 at 12:45
add a comment |Â
up vote
1
down vote
ok, you can use tangent half-angle substitution. (https://en.wikipedia.org/wiki/Tangent_half-angle_substitution)
$$int frac1(3cos(x)+2)^2 dx = int frac1left(frac3(1-tan^2(fracx2))tan^2(fracx2)+1+2right)^2dx$$
now substitute $u=tan(fracx2)$
$$Longrightarrow 2int fracu^2+1(u^2-5)^2 du$$
now you can try the new integral.
answered Aug 3 at 8:05
McBotto.t
594316
I think it is $$intfracdx(2+3cos(x))^2$$
– Dr. Sonnhard Graubner
Aug 3 at 8:08
mhhhh ok... sorry than ill remove it
– McBotto.t
Aug 3 at 8:09
add a comment |Â
up vote
1
down vote
You can use the following approach:
$$intfracdx(2+3cos x)^2=intfracdx(2cos^2fracx2+2sin^2fracx2+3cos^2fracx2-3sin^2fracx2)^2=$$
$$=intfracdx(5cos^2fracx2-sin^2fracx2)^2=intfracdxcos^4frac x2(5-tan^2frac x2)^2=left[t=tanfracx2right]=$$
$$=2intfrac1+t^2(5-t^2)^2dt=...$$
answered Aug 3 at 8:31
Mikalai Parshutsich
1097
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Integrating by parts,
$$intdfracdx(a+bcos x)^2=intdfracsin x(a+bcos x)^2cdotdfrac1sin xdx$$
$$=dfrac1sin xintdfracsin x(a+bcos x)^2dx-left(dfracd(1/sin x)dxcdotintdfracsin x(a+bcos x)^2dxright)dx$$
$$=dfrac1bsin x(a+bcos x)+intdfraccos xb(1-cos^2x)(a+bcos x),dx$$
Use Partial Fraction Decomposition,
$$dfraccos x(1-cos^2x)(a+bcos x)=dfrac A1+cos x+dfrac B1-cos x+dfrac Ca+bcos x$$
The first two integral can be managed easily, for the last use Weierstrass substitution
edited Aug 3 at 8:46
Batominovski
22.6k22776
answered Aug 3 at 8:26
lab bhattacharjee
214k14152263
add a comment |Â
up vote
2
down vote
accepted
Integrating by parts,
$$intdfracdx(a+bcos x)^2=intdfracsin x(a+bcos x)^2cdotdfrac1sin xdx$$
$$=dfrac1sin xintdfracsin x(a+bcos x)^2dx-left(dfracd(1/sin x)dxcdotintdfracsin x(a+bcos x)^2dxright)dx$$
$$=dfrac1bsin x(a+bcos x)+intdfraccos xb(1-cos^2x)(a+bcos x),dx$$
Use Partial Fraction Decomposition,
$$dfraccos x(1-cos^2x)(a+bcos x)=dfrac A1+cos x+dfrac B1-cos x+dfrac Ca+bcos x$$
The first two integral can be managed easily, for the last use Weierstrass substitution
edited Aug 3 at 8:46
Batominovski
22.6k22776
answered Aug 3 at 8:26
lab bhattacharjee
214k14152263
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Integrating by parts,
$$intdfracdx(a+bcos x)^2=intdfracsin x(a+bcos x)^2cdotdfrac1sin xdx$$
$$=dfrac1sin xintdfracsin x(a+bcos x)^2dx-left(dfracd(1/sin x)dxcdotintdfracsin x(a+bcos x)^2dxright)dx$$
$$=dfrac1bsin x(a+bcos x)+intdfraccos xb(1-cos^2x)(a+bcos x),dx$$
Use Partial Fraction Decomposition,
$$dfraccos x(1-cos^2x)(a+bcos x)=dfrac A1+cos x+dfrac B1-cos x+dfrac Ca+bcos x$$
The first two integral can be managed easily, for the last use Weierstrass substitution
edited Aug 3 at 8:46
Batominovski
22.6k22776
answered Aug 3 at 8:26
lab bhattacharjee
214k14152263
Integrating by parts,
$$intdfracdx(a+bcos x)^2=intdfracsin x(a+bcos x)^2cdotdfrac1sin xdx$$
$$=dfrac1sin xintdfracsin x(a+bcos x)^2dx-left(dfracd(1/sin x)dxcdotintdfracsin x(a+bcos x)^2dxright)dx$$
$$=dfrac1bsin x(a+bcos x)+intdfraccos xb(1-cos^2x)(a+bcos x),dx$$
Use Partial Fraction Decomposition,
$$dfraccos x(1-cos^2x)(a+bcos x)=dfrac A1+cos x+dfrac B1-cos x+dfrac Ca+bcos x$$
The first two integral can be managed easily, for the last use Weierstrass substitution
edited Aug 3 at 8:46
Batominovski
22.6k22776
answered Aug 3 at 8:26
lab bhattacharjee
214k14152263
edited Aug 3 at 8:46
Batominovski
22.6k22776
edited Aug 3 at 8:46
Batominovski
22.6k22776
edited Aug 3 at 8:46
Batominovski
22.6k22776
22.6k22776
answered Aug 3 at 8:26
lab bhattacharjee
214k14152263
answered Aug 3 at 8:26
lab bhattacharjee
214k14152263
answered Aug 3 at 8:26
lab bhattacharjee
214k14152263
214k14152263
add a comment |Â
add a comment |Â
up vote
2
down vote
Thanks guys for answering my question .
I researched on it and found out a way to solve which I liked much.
So I want share to you guys too
Please have a look .
1
The idea is beautiful, you just reversed $2+3cos x$ with $3+2cos x$ by mistake. Good job!
– Zacky
Aug 3 at 12:37
Can you please elaborate where I did it
– Ritik
Aug 3 at 12:42
1
third row, after you differentiated $A$. Nevermind, its correct. Im tired.
– Zacky
Aug 3 at 12:44
1
Sir or mam when you will differentiate that A then You will get there 2 cosx +3 that was intentional
– Ritik
Aug 3 at 12:45
add a comment |Â
up vote
2
down vote
Thanks guys for answering my question .
I researched on it and found out a way to solve which I liked much.
So I want share to you guys too
Please have a look .
1
The idea is beautiful, you just reversed $2+3cos x$ with $3+2cos x$ by mistake. Good job!
– Zacky
Aug 3 at 12:37
Can you please elaborate where I did it
– Ritik
Aug 3 at 12:42
1
third row, after you differentiated $A$. Nevermind, its correct. Im tired.
– Zacky
Aug 3 at 12:44
1
Sir or mam when you will differentiate that A then You will get there 2 cosx +3 that was intentional
– Ritik
Aug 3 at 12:45
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Thanks guys for answering my question .
I researched on it and found out a way to solve which I liked much.
So I want share to you guys too
Please have a look .
Thanks guys for answering my question .
I researched on it and found out a way to solve which I liked much.
So I want share to you guys too
Please have a look .
answered Aug 3 at 10:06
community wiki
Ritik
1
The idea is beautiful, you just reversed $2+3cos x$ with $3+2cos x$ by mistake. Good job!
– Zacky
Aug 3 at 12:37
Can you please elaborate where I did it
– Ritik
Aug 3 at 12:42
1
third row, after you differentiated $A$. Nevermind, its correct. Im tired.
– Zacky
Aug 3 at 12:44
1
Sir or mam when you will differentiate that A then You will get there 2 cosx +3 that was intentional
– Ritik
Aug 3 at 12:45
add a comment |Â
1
The idea is beautiful, you just reversed $2+3cos x$ with $3+2cos x$ by mistake. Good job!
– Zacky
Aug 3 at 12:37
Can you please elaborate where I did it
– Ritik
Aug 3 at 12:42
1
third row, after you differentiated $A$. Nevermind, its correct. Im tired.
– Zacky
Aug 3 at 12:44
1
Sir or mam when you will differentiate that A then You will get there 2 cosx +3 that was intentional
– Ritik
Aug 3 at 12:45
1
1
The idea is beautiful, you just reversed $2+3cos x$ with $3+2cos x$ by mistake. Good job!
– Zacky
Aug 3 at 12:37
The idea is beautiful, you just reversed $2+3cos x$ with $3+2cos x$ by mistake. Good job!
– Zacky
Aug 3 at 12:37
Can you please elaborate where I did it
– Ritik
Aug 3 at 12:42
Can you please elaborate where I did it
– Ritik
Aug 3 at 12:42
1
1
third row, after you differentiated $A$. Nevermind, its correct. Im tired.
– Zacky
Aug 3 at 12:44
third row, after you differentiated $A$. Nevermind, its correct. Im tired.
– Zacky
Aug 3 at 12:44
1
1
Sir or mam when you will differentiate that A then You will get there 2 cosx +3 that was intentional
– Ritik
Aug 3 at 12:45
Sir or mam when you will differentiate that A then You will get there 2 cosx +3 that was intentional
– Ritik
Aug 3 at 12:45
add a comment |Â
up vote
1
down vote
ok, you can use tangent half-angle substitution. (https://en.wikipedia.org/wiki/Tangent_half-angle_substitution)
$$int frac1(3cos(x)+2)^2 dx = int frac1left(frac3(1-tan^2(fracx2))tan^2(fracx2)+1+2right)^2dx$$
now substitute $u=tan(fracx2)$
$$Longrightarrow 2int fracu^2+1(u^2-5)^2 du$$
now you can try the new integral.
answered Aug 3 at 8:05
McBotto.t
594316
I think it is $$intfracdx(2+3cos(x))^2$$
– Dr. Sonnhard Graubner
Aug 3 at 8:08
mhhhh ok... sorry than ill remove it
– McBotto.t
Aug 3 at 8:09
add a comment |Â
up vote
1
down vote
ok, you can use tangent half-angle substitution. (https://en.wikipedia.org/wiki/Tangent_half-angle_substitution)
$$int frac1(3cos(x)+2)^2 dx = int frac1left(frac3(1-tan^2(fracx2))tan^2(fracx2)+1+2right)^2dx$$
now substitute $u=tan(fracx2)$
$$Longrightarrow 2int fracu^2+1(u^2-5)^2 du$$
now you can try the new integral.
answered Aug 3 at 8:05
McBotto.t
594316
I think it is $$intfracdx(2+3cos(x))^2$$
– Dr. Sonnhard Graubner
Aug 3 at 8:08
mhhhh ok... sorry than ill remove it
– McBotto.t
Aug 3 at 8:09
add a comment |Â
up vote
1
down vote
up vote
1
down vote
ok, you can use tangent half-angle substitution. (https://en.wikipedia.org/wiki/Tangent_half-angle_substitution)
$$int frac1(3cos(x)+2)^2 dx = int frac1left(frac3(1-tan^2(fracx2))tan^2(fracx2)+1+2right)^2dx$$
now substitute $u=tan(fracx2)$
$$Longrightarrow 2int fracu^2+1(u^2-5)^2 du$$
now you can try the new integral.
answered Aug 3 at 8:05
McBotto.t
594316
ok, you can use tangent half-angle substitution. (https://en.wikipedia.org/wiki/Tangent_half-angle_substitution)
$$int frac1(3cos(x)+2)^2 dx = int frac1left(frac3(1-tan^2(fracx2))tan^2(fracx2)+1+2right)^2dx$$
now substitute $u=tan(fracx2)$
$$Longrightarrow 2int fracu^2+1(u^2-5)^2 du$$
now you can try the new integral.
answered Aug 3 at 8:05
McBotto.t
594316
edited Aug 3 at 9:12
answered Aug 3 at 8:05
McBotto.t
594316
answered Aug 3 at 8:05
McBotto.t
594316
answered Aug 3 at 8:05
McBotto.t
594316
594316
I think it is $$intfracdx(2+3cos(x))^2$$
– Dr. Sonnhard Graubner
Aug 3 at 8:08
mhhhh ok... sorry than ill remove it
– McBotto.t
Aug 3 at 8:09
add a comment |Â
I think it is $$intfracdx(2+3cos(x))^2$$
– Dr. Sonnhard Graubner
Aug 3 at 8:08
mhhhh ok... sorry than ill remove it
– McBotto.t
Aug 3 at 8:09
I think it is $$intfracdx(2+3cos(x))^2$$
– Dr. Sonnhard Graubner
Aug 3 at 8:08
I think it is $$intfracdx(2+3cos(x))^2$$
– Dr. Sonnhard Graubner
Aug 3 at 8:08
mhhhh ok... sorry than ill remove it
– McBotto.t
Aug 3 at 8:09
mhhhh ok... sorry than ill remove it
– McBotto.t
Aug 3 at 8:09
add a comment |Â
up vote
1
down vote
You can use the following approach:
$$intfracdx(2+3cos x)^2=intfracdx(2cos^2fracx2+2sin^2fracx2+3cos^2fracx2-3sin^2fracx2)^2=$$
$$=intfracdx(5cos^2fracx2-sin^2fracx2)^2=intfracdxcos^4frac x2(5-tan^2frac x2)^2=left[t=tanfracx2right]=$$
$$=2intfrac1+t^2(5-t^2)^2dt=...$$
answered Aug 3 at 8:31
Mikalai Parshutsich
1097
add a comment |Â
up vote
1
down vote
You can use the following approach:
$$intfracdx(2+3cos x)^2=intfracdx(2cos^2fracx2+2sin^2fracx2+3cos^2fracx2-3sin^2fracx2)^2=$$
$$=intfracdx(5cos^2fracx2-sin^2fracx2)^2=intfracdxcos^4frac x2(5-tan^2frac x2)^2=left[t=tanfracx2right]=$$
$$=2intfrac1+t^2(5-t^2)^2dt=...$$
answered Aug 3 at 8:31
Mikalai Parshutsich
1097
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can use the following approach:
$$intfracdx(2+3cos x)^2=intfracdx(2cos^2fracx2+2sin^2fracx2+3cos^2fracx2-3sin^2fracx2)^2=$$
$$=intfracdx(5cos^2fracx2-sin^2fracx2)^2=intfracdxcos^4frac x2(5-tan^2frac x2)^2=left[t=tanfracx2right]=$$
$$=2intfrac1+t^2(5-t^2)^2dt=...$$
answered Aug 3 at 8:31
Mikalai Parshutsich
1097
You can use the following approach:
$$intfracdx(2+3cos x)^2=intfracdx(2cos^2fracx2+2sin^2fracx2+3cos^2fracx2-3sin^2fracx2)^2=$$
$$=intfracdx(5cos^2fracx2-sin^2fracx2)^2=intfracdxcos^4frac x2(5-tan^2frac x2)^2=left[t=tanfracx2right]=$$
$$=2intfrac1+t^2(5-t^2)^2dt=...$$
answered Aug 3 at 8:31
Mikalai Parshutsich
1097
edited Aug 3 at 9:16
answered Aug 3 at 8:31
Mikalai Parshutsich
1097
answered Aug 3 at 8:31
Mikalai Parshutsich
1097
answered Aug 3 at 8:31
Mikalai Parshutsich
1097
1097
add a comment |Â
add a comment |Â
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in.answers.yahoo.com/question/…
– lab bhattacharjee
Aug 3 at 7:41
Hint Multiply the integrand by $fracsec^2 xsec^2 x$.
– Travis
Aug 3 at 8:26