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How to calculate $int fraccos^2 x1 + sin^2 xdx$
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My approach was to use a double-arc formula for cosine and sine, but I did not succeed.
$sin^2 x = frac1 - cos(2x)2$ and $cos^2x = frac1 + cos(2x)2$
calculus integration
asked Jul 15 at 22:19
MathFacts
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up vote
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My approach was to use a double-arc formula for cosine and sine, but I did not succeed.
$sin^2 x = frac1 - cos(2x)2$ and $cos^2x = frac1 + cos(2x)2$
calculus integration
asked Jul 15 at 22:19
MathFacts
2,9921425
I would try using $1=cos^2 x+sin^2 x$ then make an obvious substitution.
– Shaun
Jul 15 at 22:23
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up vote
2
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favorite
up vote
2
down vote
favorite
My approach was to use a double-arc formula for cosine and sine, but I did not succeed.
$sin^2 x = frac1 - cos(2x)2$ and $cos^2x = frac1 + cos(2x)2$
calculus integration
asked Jul 15 at 22:19
MathFacts
2,9921425
My approach was to use a double-arc formula for cosine and sine, but I did not succeed.
$sin^2 x = frac1 - cos(2x)2$ and $cos^2x = frac1 + cos(2x)2$
calculus integration
asked Jul 15 at 22:19
MathFacts
2,9921425
edited Jul 15 at 22:50
asked Jul 15 at 22:19
MathFacts
2,9921425
asked Jul 15 at 22:19
MathFacts
2,9921425
asked Jul 15 at 22:19
MathFacts
2,9921425
2,9921425
I would try using $1=cos^2 x+sin^2 x$ then make an obvious substitution.
– Shaun
Jul 15 at 22:23
add a comment |Â
I would try using $1=cos^2 x+sin^2 x$ then make an obvious substitution.
– Shaun
Jul 15 at 22:23
I would try using $1=cos^2 x+sin^2 x$ then make an obvious substitution.
– Shaun
Jul 15 at 22:23
I would try using $1=cos^2 x+sin^2 x$ then make an obvious substitution.
– Shaun
Jul 15 at 22:23
add a comment |Â
7 Answers
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up vote
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BE CAREFUL:
$fraccos^2(x)1+sin^2(x)$ is a continuous function over $mathbbR$, therefore you would generally want an antiderivative which also has that property. It is here that WolframAlpha is evidently inferior to GeoGebra (with integration) as latter gives the wonderful solution:
$$ int fraccos^2(x)1+sin^2(x) dx = sqrt2 arctan left( frac(2-sqrt2)sin2x(sqrt2 - 2)cos2x +sqrt2 +2 right) +(sqrt2 -1)x +C$$
By graphical inspection (or otherwise), after implementing the Weierstrass substitution, one can deduce that the continuous integral is
$$ sqrt2 arctan(sqrt2 tan(x)) -x + k leftlfloor fracx+pi/2pi rightrfloor + C $$
Where, $f(x) := sqrt2 arctan(sqrt2 tan(x)) -x $,
$$k = lim_x to pi/2 ^- f(x) - lim_x to pi/2 ^+ f(x) $$
Then, to retrieve GeoGebra's beautiful closed form solution, use:
$$ lfloor x rfloor = frac1pi(arctan(cot(pi x))+pi x-pi/2)$$
Read Szeto's answer here: Continuous antiderivative of $frac11+cos^2 x$ without the floor function. for the floor function insight and concerns about discontinuity.
answered Jul 16 at 6:59
jiaminglimjm
347314
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up vote
2
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Try this
$$int dfraccos^2x1+sin^2xdx=intdfrac1-sin^2x1+sin^2xdx=int biggl(dfrac21+sin^2x-1biggr)dx$$
Can you take it from here?
answered Jul 15 at 22:22
Key Flex
4,416525
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We reduce the degree of the numerator with
$$intfraccos^2x1+sin^2xdx=intfrac2-(1+sin^2x)1+sin^2xdx=2intfracdx1+sin^2x-x$$
and we introduct a tangent to rationalize,
$$intfracdx1+sin^2x=intfracdxcos^2x+2sin^2x=intfracdxcos^2x(1+2tan^2x)=intfracdt1+2t^2
\=frac1sqrt2intfracdu1+u^2=frac1sqrt2arctan u=frac1sqrt2arctansqrt2t=frac1sqrt2arctan(sqrt2tan x).$$
answered Jul 16 at 5:40
Yves Daoust
111k665204
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up vote
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Bioche's rules suggest the substitution $; t=tan x$, so $mathrm d t=(1+t^2),mathrm dxiff mathrm dx=dfracmathrm d t1+t^2$.
Indeed, $;cos^2x=dfrac 11+t^2$, $;sin^2x=t^2cos^2x=dfrac t^21+t^2$, so that
$$int fraccos^2 x1 + sin^2 x,mathrm dx=int fracmathrm d t(1 + t^2)(1 + 2t^2)$$
There remains to decompose this fraction into partial fractions:
$$ frac1(1 + t^2)(1 + 2t^2)= fracA(1 + t^2)+fracB(1 + 2t^2).$$
Can you proceed from there?
answered Jul 15 at 23:04
Bernard
110k635103
Do you have a source for Bioche's rule? I am interested in it, but I did not find anything usefull yet. Just fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche, but I do not speak french.
– Cornman
Jul 16 at 7:03
I've found this online Google excerpt of a book by Sean M. Stewart, which explains what I learnt when I was a student.
– Bernard
Jul 16 at 7:30
add a comment |Â
up vote
1
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HINT Proceed with the following substitution $tanfracx2=t$, then $x=2arctan t$, $dx=2dt/(1+t^2)$.
Remember that $sin x=frac2t1+t^2$ and $cos x=frac1-t^21+t^2$.
answered Jul 15 at 22:25
Jack J.
3661317
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Note that we can write
$$beginalign
fraccos^2(x)1+sin^2(x)&=frac1+cos(2x)3-cos(2x)\\
&=frac4-(3-cos(2x))3-cos(2x)\\
&=-1+frac43-cos(2x)
endalign$$
Enforce the substitution $y=tan(x)$, so that $cos(2x)=frac1-y^21+y^2$ and $dx=frac11+y^2,dy$. Proceeding reveals
$$beginalign
intfrac43-cos(2x),dx&=int frac21+2y^2,dy\\
&=sqrt2arctan(sqrt2,y)+C\\
&=sqrt2arctan(sqrt2 tan(x))+C
endalign$$
Putting it all together yield
$$int fraccos^2(x)1+sin^2(x),dx=-x+sqrt2arctan(sqrt2 tan(x))+C$$
answered Jul 15 at 23:54
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
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0
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Use the substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
$$dx=frac2t1+t^2dt$$the so called Weierstrass Substitution.
answered Jul 15 at 22:25
Dr. Sonnhard Graubner
66.9k32659
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
BE CAREFUL:
$fraccos^2(x)1+sin^2(x)$ is a continuous function over $mathbbR$, therefore you would generally want an antiderivative which also has that property. It is here that WolframAlpha is evidently inferior to GeoGebra (with integration) as latter gives the wonderful solution:
$$ int fraccos^2(x)1+sin^2(x) dx = sqrt2 arctan left( frac(2-sqrt2)sin2x(sqrt2 - 2)cos2x +sqrt2 +2 right) +(sqrt2 -1)x +C$$
By graphical inspection (or otherwise), after implementing the Weierstrass substitution, one can deduce that the continuous integral is
$$ sqrt2 arctan(sqrt2 tan(x)) -x + k leftlfloor fracx+pi/2pi rightrfloor + C $$
Where, $f(x) := sqrt2 arctan(sqrt2 tan(x)) -x $,
$$k = lim_x to pi/2 ^- f(x) - lim_x to pi/2 ^+ f(x) $$
Then, to retrieve GeoGebra's beautiful closed form solution, use:
$$ lfloor x rfloor = frac1pi(arctan(cot(pi x))+pi x-pi/2)$$
Read Szeto's answer here: Continuous antiderivative of $frac11+cos^2 x$ without the floor function. for the floor function insight and concerns about discontinuity.
answered Jul 16 at 6:59
jiaminglimjm
347314
add a comment |Â
up vote
3
down vote
BE CAREFUL:
$fraccos^2(x)1+sin^2(x)$ is a continuous function over $mathbbR$, therefore you would generally want an antiderivative which also has that property. It is here that WolframAlpha is evidently inferior to GeoGebra (with integration) as latter gives the wonderful solution:
$$ int fraccos^2(x)1+sin^2(x) dx = sqrt2 arctan left( frac(2-sqrt2)sin2x(sqrt2 - 2)cos2x +sqrt2 +2 right) +(sqrt2 -1)x +C$$
By graphical inspection (or otherwise), after implementing the Weierstrass substitution, one can deduce that the continuous integral is
$$ sqrt2 arctan(sqrt2 tan(x)) -x + k leftlfloor fracx+pi/2pi rightrfloor + C $$
Where, $f(x) := sqrt2 arctan(sqrt2 tan(x)) -x $,
$$k = lim_x to pi/2 ^- f(x) - lim_x to pi/2 ^+ f(x) $$
Then, to retrieve GeoGebra's beautiful closed form solution, use:
$$ lfloor x rfloor = frac1pi(arctan(cot(pi x))+pi x-pi/2)$$
Read Szeto's answer here: Continuous antiderivative of $frac11+cos^2 x$ without the floor function. for the floor function insight and concerns about discontinuity.
answered Jul 16 at 6:59
jiaminglimjm
347314
add a comment |Â
up vote
3
down vote
up vote
3
down vote
BE CAREFUL:
$fraccos^2(x)1+sin^2(x)$ is a continuous function over $mathbbR$, therefore you would generally want an antiderivative which also has that property. It is here that WolframAlpha is evidently inferior to GeoGebra (with integration) as latter gives the wonderful solution:
$$ int fraccos^2(x)1+sin^2(x) dx = sqrt2 arctan left( frac(2-sqrt2)sin2x(sqrt2 - 2)cos2x +sqrt2 +2 right) +(sqrt2 -1)x +C$$
By graphical inspection (or otherwise), after implementing the Weierstrass substitution, one can deduce that the continuous integral is
$$ sqrt2 arctan(sqrt2 tan(x)) -x + k leftlfloor fracx+pi/2pi rightrfloor + C $$
Where, $f(x) := sqrt2 arctan(sqrt2 tan(x)) -x $,
$$k = lim_x to pi/2 ^- f(x) - lim_x to pi/2 ^+ f(x) $$
Then, to retrieve GeoGebra's beautiful closed form solution, use:
$$ lfloor x rfloor = frac1pi(arctan(cot(pi x))+pi x-pi/2)$$
Read Szeto's answer here: Continuous antiderivative of $frac11+cos^2 x$ without the floor function. for the floor function insight and concerns about discontinuity.
answered Jul 16 at 6:59
jiaminglimjm
347314
BE CAREFUL:
$fraccos^2(x)1+sin^2(x)$ is a continuous function over $mathbbR$, therefore you would generally want an antiderivative which also has that property. It is here that WolframAlpha is evidently inferior to GeoGebra (with integration) as latter gives the wonderful solution:
$$ int fraccos^2(x)1+sin^2(x) dx = sqrt2 arctan left( frac(2-sqrt2)sin2x(sqrt2 - 2)cos2x +sqrt2 +2 right) +(sqrt2 -1)x +C$$
By graphical inspection (or otherwise), after implementing the Weierstrass substitution, one can deduce that the continuous integral is
$$ sqrt2 arctan(sqrt2 tan(x)) -x + k leftlfloor fracx+pi/2pi rightrfloor + C $$
Where, $f(x) := sqrt2 arctan(sqrt2 tan(x)) -x $,
$$k = lim_x to pi/2 ^- f(x) - lim_x to pi/2 ^+ f(x) $$
Then, to retrieve GeoGebra's beautiful closed form solution, use:
$$ lfloor x rfloor = frac1pi(arctan(cot(pi x))+pi x-pi/2)$$
Read Szeto's answer here: Continuous antiderivative of $frac11+cos^2 x$ without the floor function. for the floor function insight and concerns about discontinuity.
answered Jul 16 at 6:59
jiaminglimjm
347314
answered Jul 16 at 6:59
jiaminglimjm
347314
answered Jul 16 at 6:59
jiaminglimjm
347314
answered Jul 16 at 6:59
jiaminglimjm
347314
347314
add a comment |Â
add a comment |Â
up vote
2
down vote
Try this
$$int dfraccos^2x1+sin^2xdx=intdfrac1-sin^2x1+sin^2xdx=int biggl(dfrac21+sin^2x-1biggr)dx$$
Can you take it from here?
answered Jul 15 at 22:22
Key Flex
4,416525
add a comment |Â
up vote
2
down vote
Try this
$$int dfraccos^2x1+sin^2xdx=intdfrac1-sin^2x1+sin^2xdx=int biggl(dfrac21+sin^2x-1biggr)dx$$
Can you take it from here?
answered Jul 15 at 22:22
Key Flex
4,416525
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Try this
$$int dfraccos^2x1+sin^2xdx=intdfrac1-sin^2x1+sin^2xdx=int biggl(dfrac21+sin^2x-1biggr)dx$$
Can you take it from here?
answered Jul 15 at 22:22
Key Flex
4,416525
Try this
$$int dfraccos^2x1+sin^2xdx=intdfrac1-sin^2x1+sin^2xdx=int biggl(dfrac21+sin^2x-1biggr)dx$$
Can you take it from here?
answered Jul 15 at 22:22
Key Flex
4,416525
edited Jul 15 at 22:33
answered Jul 15 at 22:22
Key Flex
4,416525
answered Jul 15 at 22:22
Key Flex
4,416525
answered Jul 15 at 22:22
Key Flex
4,416525
4,416525
add a comment |Â
add a comment |Â
up vote
2
down vote
We reduce the degree of the numerator with
$$intfraccos^2x1+sin^2xdx=intfrac2-(1+sin^2x)1+sin^2xdx=2intfracdx1+sin^2x-x$$
and we introduct a tangent to rationalize,
$$intfracdx1+sin^2x=intfracdxcos^2x+2sin^2x=intfracdxcos^2x(1+2tan^2x)=intfracdt1+2t^2
\=frac1sqrt2intfracdu1+u^2=frac1sqrt2arctan u=frac1sqrt2arctansqrt2t=frac1sqrt2arctan(sqrt2tan x).$$
answered Jul 16 at 5:40
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
We reduce the degree of the numerator with
$$intfraccos^2x1+sin^2xdx=intfrac2-(1+sin^2x)1+sin^2xdx=2intfracdx1+sin^2x-x$$
and we introduct a tangent to rationalize,
$$intfracdx1+sin^2x=intfracdxcos^2x+2sin^2x=intfracdxcos^2x(1+2tan^2x)=intfracdt1+2t^2
\=frac1sqrt2intfracdu1+u^2=frac1sqrt2arctan u=frac1sqrt2arctansqrt2t=frac1sqrt2arctan(sqrt2tan x).$$
answered Jul 16 at 5:40
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We reduce the degree of the numerator with
$$intfraccos^2x1+sin^2xdx=intfrac2-(1+sin^2x)1+sin^2xdx=2intfracdx1+sin^2x-x$$
and we introduct a tangent to rationalize,
$$intfracdx1+sin^2x=intfracdxcos^2x+2sin^2x=intfracdxcos^2x(1+2tan^2x)=intfracdt1+2t^2
\=frac1sqrt2intfracdu1+u^2=frac1sqrt2arctan u=frac1sqrt2arctansqrt2t=frac1sqrt2arctan(sqrt2tan x).$$
answered Jul 16 at 5:40
Yves Daoust
111k665204
We reduce the degree of the numerator with
$$intfraccos^2x1+sin^2xdx=intfrac2-(1+sin^2x)1+sin^2xdx=2intfracdx1+sin^2x-x$$
and we introduct a tangent to rationalize,
$$intfracdx1+sin^2x=intfracdxcos^2x+2sin^2x=intfracdxcos^2x(1+2tan^2x)=intfracdt1+2t^2
\=frac1sqrt2intfracdu1+u^2=frac1sqrt2arctan u=frac1sqrt2arctansqrt2t=frac1sqrt2arctan(sqrt2tan x).$$
answered Jul 16 at 5:40
Yves Daoust
111k665204
edited Jul 16 at 5:49
answered Jul 16 at 5:40
Yves Daoust
111k665204
answered Jul 16 at 5:40
Yves Daoust
111k665204
answered Jul 16 at 5:40
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
up vote
2
down vote
Bioche's rules suggest the substitution $; t=tan x$, so $mathrm d t=(1+t^2),mathrm dxiff mathrm dx=dfracmathrm d t1+t^2$.
Indeed, $;cos^2x=dfrac 11+t^2$, $;sin^2x=t^2cos^2x=dfrac t^21+t^2$, so that
$$int fraccos^2 x1 + sin^2 x,mathrm dx=int fracmathrm d t(1 + t^2)(1 + 2t^2)$$
There remains to decompose this fraction into partial fractions:
$$ frac1(1 + t^2)(1 + 2t^2)= fracA(1 + t^2)+fracB(1 + 2t^2).$$
Can you proceed from there?
answered Jul 15 at 23:04
Bernard
110k635103
Do you have a source for Bioche's rule? I am interested in it, but I did not find anything usefull yet. Just fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche, but I do not speak french.
– Cornman
Jul 16 at 7:03
I've found this online Google excerpt of a book by Sean M. Stewart, which explains what I learnt when I was a student.
– Bernard
Jul 16 at 7:30
add a comment |Â
up vote
2
down vote
Bioche's rules suggest the substitution $; t=tan x$, so $mathrm d t=(1+t^2),mathrm dxiff mathrm dx=dfracmathrm d t1+t^2$.
Indeed, $;cos^2x=dfrac 11+t^2$, $;sin^2x=t^2cos^2x=dfrac t^21+t^2$, so that
$$int fraccos^2 x1 + sin^2 x,mathrm dx=int fracmathrm d t(1 + t^2)(1 + 2t^2)$$
There remains to decompose this fraction into partial fractions:
$$ frac1(1 + t^2)(1 + 2t^2)= fracA(1 + t^2)+fracB(1 + 2t^2).$$
Can you proceed from there?
answered Jul 15 at 23:04
Bernard
110k635103
Do you have a source for Bioche's rule? I am interested in it, but I did not find anything usefull yet. Just fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche, but I do not speak french.
– Cornman
Jul 16 at 7:03
I've found this online Google excerpt of a book by Sean M. Stewart, which explains what I learnt when I was a student.
– Bernard
Jul 16 at 7:30
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Bioche's rules suggest the substitution $; t=tan x$, so $mathrm d t=(1+t^2),mathrm dxiff mathrm dx=dfracmathrm d t1+t^2$.
Indeed, $;cos^2x=dfrac 11+t^2$, $;sin^2x=t^2cos^2x=dfrac t^21+t^2$, so that
$$int fraccos^2 x1 + sin^2 x,mathrm dx=int fracmathrm d t(1 + t^2)(1 + 2t^2)$$
There remains to decompose this fraction into partial fractions:
$$ frac1(1 + t^2)(1 + 2t^2)= fracA(1 + t^2)+fracB(1 + 2t^2).$$
Can you proceed from there?
answered Jul 15 at 23:04
Bernard
110k635103
Bioche's rules suggest the substitution $; t=tan x$, so $mathrm d t=(1+t^2),mathrm dxiff mathrm dx=dfracmathrm d t1+t^2$.
Indeed, $;cos^2x=dfrac 11+t^2$, $;sin^2x=t^2cos^2x=dfrac t^21+t^2$, so that
$$int fraccos^2 x1 + sin^2 x,mathrm dx=int fracmathrm d t(1 + t^2)(1 + 2t^2)$$
There remains to decompose this fraction into partial fractions:
$$ frac1(1 + t^2)(1 + 2t^2)= fracA(1 + t^2)+fracB(1 + 2t^2).$$
Can you proceed from there?
answered Jul 15 at 23:04
Bernard
110k635103
edited Jul 16 at 7:14
answered Jul 15 at 23:04
Bernard
110k635103
answered Jul 15 at 23:04
Bernard
110k635103
answered Jul 15 at 23:04
Bernard
110k635103
110k635103
Do you have a source for Bioche's rule? I am interested in it, but I did not find anything usefull yet. Just fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche, but I do not speak french.
– Cornman
Jul 16 at 7:03
I've found this online Google excerpt of a book by Sean M. Stewart, which explains what I learnt when I was a student.
– Bernard
Jul 16 at 7:30
add a comment |Â
Do you have a source for Bioche's rule? I am interested in it, but I did not find anything usefull yet. Just fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche, but I do not speak french.
– Cornman
Jul 16 at 7:03
I've found this online Google excerpt of a book by Sean M. Stewart, which explains what I learnt when I was a student.
– Bernard
Jul 16 at 7:30
Do you have a source for Bioche's rule? I am interested in it, but I did not find anything usefull yet. Just fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche, but I do not speak french.
– Cornman
Jul 16 at 7:03
Do you have a source for Bioche's rule? I am interested in it, but I did not find anything usefull yet. Just fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche, but I do not speak french.
– Cornman
Jul 16 at 7:03
I've found this online Google excerpt of a book by Sean M. Stewart, which explains what I learnt when I was a student.
– Bernard
Jul 16 at 7:30
I've found this online Google excerpt of a book by Sean M. Stewart, which explains what I learnt when I was a student.
– Bernard
Jul 16 at 7:30
add a comment |Â
up vote
1
down vote
HINT Proceed with the following substitution $tanfracx2=t$, then $x=2arctan t$, $dx=2dt/(1+t^2)$.
Remember that $sin x=frac2t1+t^2$ and $cos x=frac1-t^21+t^2$.
answered Jul 15 at 22:25
Jack J.
3661317
add a comment |Â
up vote
1
down vote
HINT Proceed with the following substitution $tanfracx2=t$, then $x=2arctan t$, $dx=2dt/(1+t^2)$.
Remember that $sin x=frac2t1+t^2$ and $cos x=frac1-t^21+t^2$.
answered Jul 15 at 22:25
Jack J.
3661317
add a comment |Â
up vote
1
down vote
up vote
1
down vote
HINT Proceed with the following substitution $tanfracx2=t$, then $x=2arctan t$, $dx=2dt/(1+t^2)$.
Remember that $sin x=frac2t1+t^2$ and $cos x=frac1-t^21+t^2$.
answered Jul 15 at 22:25
Jack J.
3661317
HINT Proceed with the following substitution $tanfracx2=t$, then $x=2arctan t$, $dx=2dt/(1+t^2)$.
Remember that $sin x=frac2t1+t^2$ and $cos x=frac1-t^21+t^2$.
answered Jul 15 at 22:25
Jack J.
3661317
answered Jul 15 at 22:25
Jack J.
3661317
answered Jul 15 at 22:25
Jack J.
3661317
answered Jul 15 at 22:25
Jack J.
3661317
3661317
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that we can write
$$beginalign
fraccos^2(x)1+sin^2(x)&=frac1+cos(2x)3-cos(2x)\\
&=frac4-(3-cos(2x))3-cos(2x)\\
&=-1+frac43-cos(2x)
endalign$$
Enforce the substitution $y=tan(x)$, so that $cos(2x)=frac1-y^21+y^2$ and $dx=frac11+y^2,dy$. Proceeding reveals
$$beginalign
intfrac43-cos(2x),dx&=int frac21+2y^2,dy\\
&=sqrt2arctan(sqrt2,y)+C\\
&=sqrt2arctan(sqrt2 tan(x))+C
endalign$$
Putting it all together yield
$$int fraccos^2(x)1+sin^2(x),dx=-x+sqrt2arctan(sqrt2 tan(x))+C$$
answered Jul 15 at 23:54
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
add a comment |Â
up vote
1
down vote
Note that we can write
$$beginalign
fraccos^2(x)1+sin^2(x)&=frac1+cos(2x)3-cos(2x)\\
&=frac4-(3-cos(2x))3-cos(2x)\\
&=-1+frac43-cos(2x)
endalign$$
Enforce the substitution $y=tan(x)$, so that $cos(2x)=frac1-y^21+y^2$ and $dx=frac11+y^2,dy$. Proceeding reveals
$$beginalign
intfrac43-cos(2x),dx&=int frac21+2y^2,dy\\
&=sqrt2arctan(sqrt2,y)+C\\
&=sqrt2arctan(sqrt2 tan(x))+C
endalign$$
Putting it all together yield
$$int fraccos^2(x)1+sin^2(x),dx=-x+sqrt2arctan(sqrt2 tan(x))+C$$
answered Jul 15 at 23:54
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
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up vote
1
down vote
up vote
1
down vote
Note that we can write
$$beginalign
fraccos^2(x)1+sin^2(x)&=frac1+cos(2x)3-cos(2x)\\
&=frac4-(3-cos(2x))3-cos(2x)\\
&=-1+frac43-cos(2x)
endalign$$
Enforce the substitution $y=tan(x)$, so that $cos(2x)=frac1-y^21+y^2$ and $dx=frac11+y^2,dy$. Proceeding reveals
$$beginalign
intfrac43-cos(2x),dx&=int frac21+2y^2,dy\\
&=sqrt2arctan(sqrt2,y)+C\\
&=sqrt2arctan(sqrt2 tan(x))+C
endalign$$
Putting it all together yield
$$int fraccos^2(x)1+sin^2(x),dx=-x+sqrt2arctan(sqrt2 tan(x))+C$$
answered Jul 15 at 23:54
Mark Viola
126k1172167
Note that we can write
$$beginalign
fraccos^2(x)1+sin^2(x)&=frac1+cos(2x)3-cos(2x)\\
&=frac4-(3-cos(2x))3-cos(2x)\\
&=-1+frac43-cos(2x)
endalign$$
Enforce the substitution $y=tan(x)$, so that $cos(2x)=frac1-y^21+y^2$ and $dx=frac11+y^2,dy$. Proceeding reveals
$$beginalign
intfrac43-cos(2x),dx&=int frac21+2y^2,dy\\
&=sqrt2arctan(sqrt2,y)+C\\
&=sqrt2arctan(sqrt2 tan(x))+C
endalign$$
Putting it all together yield
$$int fraccos^2(x)1+sin^2(x),dx=-x+sqrt2arctan(sqrt2 tan(x))+C$$
answered Jul 15 at 23:54
Mark Viola
126k1172167
answered Jul 15 at 23:54
Mark Viola
126k1172167
answered Jul 15 at 23:54
Mark Viola
126k1172167
answered Jul 15 at 23:54
Mark Viola
126k1172167
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
add a comment |Â
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
add a comment |Â
up vote
0
down vote
Use the substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
$$dx=frac2t1+t^2dt$$the so called Weierstrass Substitution.
answered Jul 15 at 22:25
Dr. Sonnhard Graubner
66.9k32659
add a comment |Â
up vote
0
down vote
Use the substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
$$dx=frac2t1+t^2dt$$the so called Weierstrass Substitution.
answered Jul 15 at 22:25
Dr. Sonnhard Graubner
66.9k32659
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use the substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
$$dx=frac2t1+t^2dt$$the so called Weierstrass Substitution.
answered Jul 15 at 22:25
Dr. Sonnhard Graubner
66.9k32659
Use the substitution:
$$sin(x)=frac2t1+t^2$$
$$cos(x)=frac1-t^21+t^2$$
$$dx=frac2t1+t^2dt$$the so called Weierstrass Substitution.
answered Jul 15 at 22:25
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 22:25
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 22:25
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 22:25
Dr. Sonnhard Graubner
66.9k32659
66.9k32659
add a comment |Â
add a comment |Â
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I would try using $1=cos^2 x+sin^2 x$ then make an obvious substitution.
– Shaun
Jul 15 at 22:23