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How to integrate $int frac1sin^4x + cos^4 x ,dx$?
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How to integrate
$$int frac1sin^4x + cos^4 x ,dx$$
I tried the following approach:
$$int frac1sin^4x + cos^4 x ,dx = int frac1sin^4x + (1-sin^2x)^2 ,dx = int frac1sin^4x + 1- 2sin^2x + sin^4x ,dx \
= frac12int frac1sin^4x - sin^2x + frac12 ,dx = frac12int frac1(sin^2x - frac12)^2 + frac14 ,dx$$
The substitution $t = tanfracx2$ yields 4th degree polynomials and a $sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$sin^4x + cos^4 x = (sin^2 x + cos^2x)(sin^2 x + cos^2 x) - 2sin^2 xcos^2 x = 1 - 2sin^2 xcos^2 x = (1-sqrt2sin x cos x)(1+sqrt2sin x cos x)$
and then I tried substituting: $t = sin x cos x$ and got
$$intfract,dt2(1-2t^2)sqrt1-4t^2$$
Another way would maybe be to make two integrals:
$$int frac1sin^4x + cos^4 x ,dx = int frac1(1-sqrt2sin x cos x)(1+sqrt2sin x cos x) ,dx = \ frac12int frac11-sqrt2sin x cos x ,dx + frac12intfrac11+sqrt2sin x cos x ,dx$$
... and again I tried $t = tanfracx2$ (4th degree polynomial) and $t=sqrt2 sin x cos x$ and I get $fracsqrt 22 int frac,dt(1-t)sqrt1-2t^2$ for the first one.
Any hints?
calculus real-analysis integration indefinite-integrals
edited Jun 5 '14 at 8:15
Tunk-Fey
22.7k866100
asked Jun 4 '14 at 19:40
AltairAC
5391418
add a comment |Â
up vote
11
down vote
favorite
How to integrate
$$int frac1sin^4x + cos^4 x ,dx$$
I tried the following approach:
$$int frac1sin^4x + cos^4 x ,dx = int frac1sin^4x + (1-sin^2x)^2 ,dx = int frac1sin^4x + 1- 2sin^2x + sin^4x ,dx \
= frac12int frac1sin^4x - sin^2x + frac12 ,dx = frac12int frac1(sin^2x - frac12)^2 + frac14 ,dx$$
The substitution $t = tanfracx2$ yields 4th degree polynomials and a $sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$sin^4x + cos^4 x = (sin^2 x + cos^2x)(sin^2 x + cos^2 x) - 2sin^2 xcos^2 x = 1 - 2sin^2 xcos^2 x = (1-sqrt2sin x cos x)(1+sqrt2sin x cos x)$
and then I tried substituting: $t = sin x cos x$ and got
$$intfract,dt2(1-2t^2)sqrt1-4t^2$$
Another way would maybe be to make two integrals:
$$int frac1sin^4x + cos^4 x ,dx = int frac1(1-sqrt2sin x cos x)(1+sqrt2sin x cos x) ,dx = \ frac12int frac11-sqrt2sin x cos x ,dx + frac12intfrac11+sqrt2sin x cos x ,dx$$
... and again I tried $t = tanfracx2$ (4th degree polynomial) and $t=sqrt2 sin x cos x$ and I get $fracsqrt 22 int frac,dt(1-t)sqrt1-2t^2$ for the first one.
Any hints?
calculus real-analysis integration indefinite-integrals
edited Jun 5 '14 at 8:15
Tunk-Fey
22.7k866100
asked Jun 4 '14 at 19:40
AltairAC
5391418
Have you tried the Weierstrass substitution?
– Lucian
Jun 4 '14 at 19:44
Using the en.wikipedia.org/wiki/… will transform the integrand to the simpler $frac1a+bcos4t$.
– Yves Daoust
Jun 4 '14 at 19:48
8
Notice $$sin^4 x + cos^4x = 1 - 2sin^2xcos^2x = 1 - frac12sin^2(2x) = frac14(3+cos4x)$$ Introduce $t = tan2x$, we get $$intfracdxsin^4 x + cos^4x = int frac43 + frac1-t^21+t^2fracdt2(1+t^2) = intfracdt2+t^2 = frac1sqrt2tan^-1left(fractsqrt2right) + textconst.$$
– achille hui
Jun 4 '14 at 19:55
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
How to integrate
$$int frac1sin^4x + cos^4 x ,dx$$
I tried the following approach:
$$int frac1sin^4x + cos^4 x ,dx = int frac1sin^4x + (1-sin^2x)^2 ,dx = int frac1sin^4x + 1- 2sin^2x + sin^4x ,dx \
= frac12int frac1sin^4x - sin^2x + frac12 ,dx = frac12int frac1(sin^2x - frac12)^2 + frac14 ,dx$$
The substitution $t = tanfracx2$ yields 4th degree polynomials and a $sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$sin^4x + cos^4 x = (sin^2 x + cos^2x)(sin^2 x + cos^2 x) - 2sin^2 xcos^2 x = 1 - 2sin^2 xcos^2 x = (1-sqrt2sin x cos x)(1+sqrt2sin x cos x)$
and then I tried substituting: $t = sin x cos x$ and got
$$intfract,dt2(1-2t^2)sqrt1-4t^2$$
Another way would maybe be to make two integrals:
$$int frac1sin^4x + cos^4 x ,dx = int frac1(1-sqrt2sin x cos x)(1+sqrt2sin x cos x) ,dx = \ frac12int frac11-sqrt2sin x cos x ,dx + frac12intfrac11+sqrt2sin x cos x ,dx$$
... and again I tried $t = tanfracx2$ (4th degree polynomial) and $t=sqrt2 sin x cos x$ and I get $fracsqrt 22 int frac,dt(1-t)sqrt1-2t^2$ for the first one.
Any hints?
calculus real-analysis integration indefinite-integrals
edited Jun 5 '14 at 8:15
Tunk-Fey
22.7k866100
asked Jun 4 '14 at 19:40
AltairAC
5391418
How to integrate
$$int frac1sin^4x + cos^4 x ,dx$$
I tried the following approach:
$$int frac1sin^4x + cos^4 x ,dx = int frac1sin^4x + (1-sin^2x)^2 ,dx = int frac1sin^4x + 1- 2sin^2x + sin^4x ,dx \
= frac12int frac1sin^4x - sin^2x + frac12 ,dx = frac12int frac1(sin^2x - frac12)^2 + frac14 ,dx$$
The substitution $t = tanfracx2$ yields 4th degree polynomials and a $sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$sin^4x + cos^4 x = (sin^2 x + cos^2x)(sin^2 x + cos^2 x) - 2sin^2 xcos^2 x = 1 - 2sin^2 xcos^2 x = (1-sqrt2sin x cos x)(1+sqrt2sin x cos x)$
and then I tried substituting: $t = sin x cos x$ and got
$$intfract,dt2(1-2t^2)sqrt1-4t^2$$
Another way would maybe be to make two integrals:
$$int frac1sin^4x + cos^4 x ,dx = int frac1(1-sqrt2sin x cos x)(1+sqrt2sin x cos x) ,dx = \ frac12int frac11-sqrt2sin x cos x ,dx + frac12intfrac11+sqrt2sin x cos x ,dx$$
... and again I tried $t = tanfracx2$ (4th degree polynomial) and $t=sqrt2 sin x cos x$ and I get $fracsqrt 22 int frac,dt(1-t)sqrt1-2t^2$ for the first one.
Any hints?
calculus real-analysis integration indefinite-integrals
edited Jun 5 '14 at 8:15
Tunk-Fey
22.7k866100
asked Jun 4 '14 at 19:40
AltairAC
5391418
edited Jun 5 '14 at 8:15
Tunk-Fey
22.7k866100
edited Jun 5 '14 at 8:15
Tunk-Fey
22.7k866100
edited Jun 5 '14 at 8:15
Tunk-Fey
22.7k866100
22.7k866100
asked Jun 4 '14 at 19:40
AltairAC
5391418
asked Jun 4 '14 at 19:40
AltairAC
5391418
asked Jun 4 '14 at 19:40
AltairAC
5391418
5391418
Have you tried the Weierstrass substitution?
– Lucian
Jun 4 '14 at 19:44
Using the en.wikipedia.org/wiki/… will transform the integrand to the simpler $frac1a+bcos4t$.
– Yves Daoust
Jun 4 '14 at 19:48
8
Notice $$sin^4 x + cos^4x = 1 - 2sin^2xcos^2x = 1 - frac12sin^2(2x) = frac14(3+cos4x)$$ Introduce $t = tan2x$, we get $$intfracdxsin^4 x + cos^4x = int frac43 + frac1-t^21+t^2fracdt2(1+t^2) = intfracdt2+t^2 = frac1sqrt2tan^-1left(fractsqrt2right) + textconst.$$
– achille hui
Jun 4 '14 at 19:55
add a comment |Â
Have you tried the Weierstrass substitution?
– Lucian
Jun 4 '14 at 19:44
Using the en.wikipedia.org/wiki/… will transform the integrand to the simpler $frac1a+bcos4t$.
– Yves Daoust
Jun 4 '14 at 19:48
8
Notice $$sin^4 x + cos^4x = 1 - 2sin^2xcos^2x = 1 - frac12sin^2(2x) = frac14(3+cos4x)$$ Introduce $t = tan2x$, we get $$intfracdxsin^4 x + cos^4x = int frac43 + frac1-t^21+t^2fracdt2(1+t^2) = intfracdt2+t^2 = frac1sqrt2tan^-1left(fractsqrt2right) + textconst.$$
– achille hui
Jun 4 '14 at 19:55
Have you tried the Weierstrass substitution?
– Lucian
Jun 4 '14 at 19:44
Have you tried the Weierstrass substitution?
– Lucian
Jun 4 '14 at 19:44
Using the en.wikipedia.org/wiki/… will transform the integrand to the simpler $frac1a+bcos4t$.
– Yves Daoust
Jun 4 '14 at 19:48
Using the en.wikipedia.org/wiki/… will transform the integrand to the simpler $frac1a+bcos4t$.
– Yves Daoust
Jun 4 '14 at 19:48
8
8
Notice $$sin^4 x + cos^4x = 1 - 2sin^2xcos^2x = 1 - frac12sin^2(2x) = frac14(3+cos4x)$$ Introduce $t = tan2x$, we get $$intfracdxsin^4 x + cos^4x = int frac43 + frac1-t^21+t^2fracdt2(1+t^2) = intfracdt2+t^2 = frac1sqrt2tan^-1left(fractsqrt2right) + textconst.$$
– achille hui
Jun 4 '14 at 19:55
Notice $$sin^4 x + cos^4x = 1 - 2sin^2xcos^2x = 1 - frac12sin^2(2x) = frac14(3+cos4x)$$ Introduce $t = tan2x$, we get $$intfracdxsin^4 x + cos^4x = int frac43 + frac1-t^21+t^2fracdt2(1+t^2) = intfracdt2+t^2 = frac1sqrt2tan^-1left(fractsqrt2right) + textconst.$$
– achille hui
Jun 4 '14 at 19:55
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
16
down vote
accepted
Simplify the denominator in the following way:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-fracsin^2(2x)2=frac1+cos^2(2x)2=frac2+tan^2(2x)2sec^2(2x)$$
Hence, the integral you are dealing with is:
$$int frac2sec^2(2x)2+tan^2(2x),dx$$
I guess the next step is pretty obvious now. ;)
answered Jun 4 '14 at 19:58
Pranav Arora
8,4822462
May I ask how you got: $frac2+tan^2(2x)2sec^2(2x)$ ? Did you use Wolfram Alpha, "try and error", a certain trick, a specific algorithm, ... ? I verified for myself that this is true but how to "come up" with something like this?
– AltairAC
Jun 13 '14 at 9:04
1
@Shirohige: I used only the basic trigonometric identities. First factor out $cos^2x$ from the numerator and use that $cos^2x=frac1sec^2x$. Then write $sec^2x$ in the numerator as $1+tan^2x$.
– Pranav Arora
Jun 13 '14 at 14:01
add a comment |Â
up vote
6
down vote
Another approach:
We have
$$
frac1sin^4x+cos^4x,tag1
$$
Multiply $(1)$ by $dfractan^4xtan^4x$ we obtain
$$
fractan^4xsin^4x(1+tan^4x)=fracsec^4x1+tan^4x=frac(1+tan^2x)sec^2x1+tan^4x.tag2
$$
Letting $t=tan x$, the integral turns out to be
$$eqalign
intfrac1+t^21+t^4 dt&=frac12intleft[frac1t^2-sqrt2t+1+frac1t^2+sqrt2t+1right] dt\
&=frac12intleft[frac1left(t-dfrac1sqrt2right)^2+dfrac34+frac1left(t+dfrac1sqrt2right)^2+dfrac34right] dt.tag3
$$
Using substitution $u=dfracsqrt32left(t-dfrac1sqrt2right)$ and $v=dfracsqrt32left(t+dfrac1sqrt2right)$, the integral in $(3)$ can easily be evaluated.
Addendum :
Another way to evaluate $displaystyleintfrac1+t^21+t^4 dt$ is dividing the integrand by $dfract^2t^2$, we obtain
$$eqalign
intfrac1+dfrac1t^2t^2+dfrac1t^2 dt&=intfrac1+dfrac1t^2left(t-dfrac1tright)^2+2 dt.
$$
Now let $u=t-dfrac1t;Rightarrow;du=left(1+dfrac1t^2right) dt$, the integral turns out to be
$$
intfrac1u^2+2 du.tag4
$$
The evaluation of the integral $(4)$ can follow @achillehui's comment.
answered Jun 5 '14 at 8:08
Tunk-Fey
22.7k866100
add a comment |Â
up vote
0
down vote
Convert the exponential powers to multiple angles. From deMoivre's theorem, with $ninmathbbN$:
$$
beginalign
left( e^i theta right)^n &= e^i ntheta \
left( cos theta + i sin theta right)^n &= cos ntheta + i sin ntheta
endalign
$$
These intermediate formulas may help:
$$
beginalign
cos 2theta &= cos^2 theta + sin^2 theta \
sin 2theta &= 2 cos theta sin 2 theta \
endalign
$$
Reduce the denominator
$$
sin ^4(x)+cos ^4(x) = frac14 (cos (4 x)+3)
$$
The primitive is
$$
int frac1cos^4x + sin^4x , dx =
int frac1cos (4 x)+3 , dx =
left(4 sqrt2right)^-1arctan left(fractan (2 x)sqrt2right)
$$
Here is a look at the integrand:
answered Apr 17 '17 at 16:17
dantopa
6,02131640
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
16
down vote
accepted
Simplify the denominator in the following way:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-fracsin^2(2x)2=frac1+cos^2(2x)2=frac2+tan^2(2x)2sec^2(2x)$$
Hence, the integral you are dealing with is:
$$int frac2sec^2(2x)2+tan^2(2x),dx$$
I guess the next step is pretty obvious now. ;)
answered Jun 4 '14 at 19:58
Pranav Arora
8,4822462
May I ask how you got: $frac2+tan^2(2x)2sec^2(2x)$ ? Did you use Wolfram Alpha, "try and error", a certain trick, a specific algorithm, ... ? I verified for myself that this is true but how to "come up" with something like this?
– AltairAC
Jun 13 '14 at 9:04
1
@Shirohige: I used only the basic trigonometric identities. First factor out $cos^2x$ from the numerator and use that $cos^2x=frac1sec^2x$. Then write $sec^2x$ in the numerator as $1+tan^2x$.
– Pranav Arora
Jun 13 '14 at 14:01
add a comment |Â
up vote
16
down vote
accepted
Simplify the denominator in the following way:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-fracsin^2(2x)2=frac1+cos^2(2x)2=frac2+tan^2(2x)2sec^2(2x)$$
Hence, the integral you are dealing with is:
$$int frac2sec^2(2x)2+tan^2(2x),dx$$
I guess the next step is pretty obvious now. ;)
answered Jun 4 '14 at 19:58
Pranav Arora
8,4822462
May I ask how you got: $frac2+tan^2(2x)2sec^2(2x)$ ? Did you use Wolfram Alpha, "try and error", a certain trick, a specific algorithm, ... ? I verified for myself that this is true but how to "come up" with something like this?
– AltairAC
Jun 13 '14 at 9:04
1
@Shirohige: I used only the basic trigonometric identities. First factor out $cos^2x$ from the numerator and use that $cos^2x=frac1sec^2x$. Then write $sec^2x$ in the numerator as $1+tan^2x$.
– Pranav Arora
Jun 13 '14 at 14:01
add a comment |Â
up vote
16
down vote
accepted
up vote
16
down vote
accepted
Simplify the denominator in the following way:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-fracsin^2(2x)2=frac1+cos^2(2x)2=frac2+tan^2(2x)2sec^2(2x)$$
Hence, the integral you are dealing with is:
$$int frac2sec^2(2x)2+tan^2(2x),dx$$
I guess the next step is pretty obvious now. ;)
answered Jun 4 '14 at 19:58
Pranav Arora
8,4822462
Simplify the denominator in the following way:
$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-fracsin^2(2x)2=frac1+cos^2(2x)2=frac2+tan^2(2x)2sec^2(2x)$$
Hence, the integral you are dealing with is:
$$int frac2sec^2(2x)2+tan^2(2x),dx$$
I guess the next step is pretty obvious now. ;)
answered Jun 4 '14 at 19:58
Pranav Arora
8,4822462
answered Jun 4 '14 at 19:58
Pranav Arora
8,4822462
answered Jun 4 '14 at 19:58
Pranav Arora
8,4822462
answered Jun 4 '14 at 19:58
Pranav Arora
8,4822462
8,4822462
May I ask how you got: $frac2+tan^2(2x)2sec^2(2x)$ ? Did you use Wolfram Alpha, "try and error", a certain trick, a specific algorithm, ... ? I verified for myself that this is true but how to "come up" with something like this?
– AltairAC
Jun 13 '14 at 9:04
1
@Shirohige: I used only the basic trigonometric identities. First factor out $cos^2x$ from the numerator and use that $cos^2x=frac1sec^2x$. Then write $sec^2x$ in the numerator as $1+tan^2x$.
– Pranav Arora
Jun 13 '14 at 14:01
add a comment |Â
May I ask how you got: $frac2+tan^2(2x)2sec^2(2x)$ ? Did you use Wolfram Alpha, "try and error", a certain trick, a specific algorithm, ... ? I verified for myself that this is true but how to "come up" with something like this?
– AltairAC
Jun 13 '14 at 9:04
1
@Shirohige: I used only the basic trigonometric identities. First factor out $cos^2x$ from the numerator and use that $cos^2x=frac1sec^2x$. Then write $sec^2x$ in the numerator as $1+tan^2x$.
– Pranav Arora
Jun 13 '14 at 14:01
May I ask how you got: $frac2+tan^2(2x)2sec^2(2x)$ ? Did you use Wolfram Alpha, "try and error", a certain trick, a specific algorithm, ... ? I verified for myself that this is true but how to "come up" with something like this?
– AltairAC
Jun 13 '14 at 9:04
May I ask how you got: $frac2+tan^2(2x)2sec^2(2x)$ ? Did you use Wolfram Alpha, "try and error", a certain trick, a specific algorithm, ... ? I verified for myself that this is true but how to "come up" with something like this?
– AltairAC
Jun 13 '14 at 9:04
1
1
@Shirohige: I used only the basic trigonometric identities. First factor out $cos^2x$ from the numerator and use that $cos^2x=frac1sec^2x$. Then write $sec^2x$ in the numerator as $1+tan^2x$.
– Pranav Arora
Jun 13 '14 at 14:01
@Shirohige: I used only the basic trigonometric identities. First factor out $cos^2x$ from the numerator and use that $cos^2x=frac1sec^2x$. Then write $sec^2x$ in the numerator as $1+tan^2x$.
– Pranav Arora
Jun 13 '14 at 14:01
add a comment |Â
up vote
6
down vote
Another approach:
We have
$$
frac1sin^4x+cos^4x,tag1
$$
Multiply $(1)$ by $dfractan^4xtan^4x$ we obtain
$$
fractan^4xsin^4x(1+tan^4x)=fracsec^4x1+tan^4x=frac(1+tan^2x)sec^2x1+tan^4x.tag2
$$
Letting $t=tan x$, the integral turns out to be
$$eqalign
intfrac1+t^21+t^4 dt&=frac12intleft[frac1t^2-sqrt2t+1+frac1t^2+sqrt2t+1right] dt\
&=frac12intleft[frac1left(t-dfrac1sqrt2right)^2+dfrac34+frac1left(t+dfrac1sqrt2right)^2+dfrac34right] dt.tag3
$$
Using substitution $u=dfracsqrt32left(t-dfrac1sqrt2right)$ and $v=dfracsqrt32left(t+dfrac1sqrt2right)$, the integral in $(3)$ can easily be evaluated.
Addendum :
Another way to evaluate $displaystyleintfrac1+t^21+t^4 dt$ is dividing the integrand by $dfract^2t^2$, we obtain
$$eqalign
intfrac1+dfrac1t^2t^2+dfrac1t^2 dt&=intfrac1+dfrac1t^2left(t-dfrac1tright)^2+2 dt.
$$
Now let $u=t-dfrac1t;Rightarrow;du=left(1+dfrac1t^2right) dt$, the integral turns out to be
$$
intfrac1u^2+2 du.tag4
$$
The evaluation of the integral $(4)$ can follow @achillehui's comment.
answered Jun 5 '14 at 8:08
Tunk-Fey
22.7k866100
add a comment |Â
up vote
6
down vote
Another approach:
We have
$$
frac1sin^4x+cos^4x,tag1
$$
Multiply $(1)$ by $dfractan^4xtan^4x$ we obtain
$$
fractan^4xsin^4x(1+tan^4x)=fracsec^4x1+tan^4x=frac(1+tan^2x)sec^2x1+tan^4x.tag2
$$
Letting $t=tan x$, the integral turns out to be
$$eqalign
intfrac1+t^21+t^4 dt&=frac12intleft[frac1t^2-sqrt2t+1+frac1t^2+sqrt2t+1right] dt\
&=frac12intleft[frac1left(t-dfrac1sqrt2right)^2+dfrac34+frac1left(t+dfrac1sqrt2right)^2+dfrac34right] dt.tag3
$$
Using substitution $u=dfracsqrt32left(t-dfrac1sqrt2right)$ and $v=dfracsqrt32left(t+dfrac1sqrt2right)$, the integral in $(3)$ can easily be evaluated.
Addendum :
Another way to evaluate $displaystyleintfrac1+t^21+t^4 dt$ is dividing the integrand by $dfract^2t^2$, we obtain
$$eqalign
intfrac1+dfrac1t^2t^2+dfrac1t^2 dt&=intfrac1+dfrac1t^2left(t-dfrac1tright)^2+2 dt.
$$
Now let $u=t-dfrac1t;Rightarrow;du=left(1+dfrac1t^2right) dt$, the integral turns out to be
$$
intfrac1u^2+2 du.tag4
$$
The evaluation of the integral $(4)$ can follow @achillehui's comment.
answered Jun 5 '14 at 8:08
Tunk-Fey
22.7k866100
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Another approach:
We have
$$
frac1sin^4x+cos^4x,tag1
$$
Multiply $(1)$ by $dfractan^4xtan^4x$ we obtain
$$
fractan^4xsin^4x(1+tan^4x)=fracsec^4x1+tan^4x=frac(1+tan^2x)sec^2x1+tan^4x.tag2
$$
Letting $t=tan x$, the integral turns out to be
$$eqalign
intfrac1+t^21+t^4 dt&=frac12intleft[frac1t^2-sqrt2t+1+frac1t^2+sqrt2t+1right] dt\
&=frac12intleft[frac1left(t-dfrac1sqrt2right)^2+dfrac34+frac1left(t+dfrac1sqrt2right)^2+dfrac34right] dt.tag3
$$
Using substitution $u=dfracsqrt32left(t-dfrac1sqrt2right)$ and $v=dfracsqrt32left(t+dfrac1sqrt2right)$, the integral in $(3)$ can easily be evaluated.
Addendum :
Another way to evaluate $displaystyleintfrac1+t^21+t^4 dt$ is dividing the integrand by $dfract^2t^2$, we obtain
$$eqalign
intfrac1+dfrac1t^2t^2+dfrac1t^2 dt&=intfrac1+dfrac1t^2left(t-dfrac1tright)^2+2 dt.
$$
Now let $u=t-dfrac1t;Rightarrow;du=left(1+dfrac1t^2right) dt$, the integral turns out to be
$$
intfrac1u^2+2 du.tag4
$$
The evaluation of the integral $(4)$ can follow @achillehui's comment.
answered Jun 5 '14 at 8:08
Tunk-Fey
22.7k866100
Another approach:
We have
$$
frac1sin^4x+cos^4x,tag1
$$
Multiply $(1)$ by $dfractan^4xtan^4x$ we obtain
$$
fractan^4xsin^4x(1+tan^4x)=fracsec^4x1+tan^4x=frac(1+tan^2x)sec^2x1+tan^4x.tag2
$$
Letting $t=tan x$, the integral turns out to be
$$eqalign
intfrac1+t^21+t^4 dt&=frac12intleft[frac1t^2-sqrt2t+1+frac1t^2+sqrt2t+1right] dt\
&=frac12intleft[frac1left(t-dfrac1sqrt2right)^2+dfrac34+frac1left(t+dfrac1sqrt2right)^2+dfrac34right] dt.tag3
$$
Using substitution $u=dfracsqrt32left(t-dfrac1sqrt2right)$ and $v=dfracsqrt32left(t+dfrac1sqrt2right)$, the integral in $(3)$ can easily be evaluated.
Addendum :
Another way to evaluate $displaystyleintfrac1+t^21+t^4 dt$ is dividing the integrand by $dfract^2t^2$, we obtain
$$eqalign
intfrac1+dfrac1t^2t^2+dfrac1t^2 dt&=intfrac1+dfrac1t^2left(t-dfrac1tright)^2+2 dt.
$$
Now let $u=t-dfrac1t;Rightarrow;du=left(1+dfrac1t^2right) dt$, the integral turns out to be
$$
intfrac1u^2+2 du.tag4
$$
The evaluation of the integral $(4)$ can follow @achillehui's comment.
answered Jun 5 '14 at 8:08
Tunk-Fey
22.7k866100
edited Jun 5 '14 at 11:21
answered Jun 5 '14 at 8:08
Tunk-Fey
22.7k866100
answered Jun 5 '14 at 8:08
Tunk-Fey
22.7k866100
answered Jun 5 '14 at 8:08
Tunk-Fey
22.7k866100
22.7k866100
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Convert the exponential powers to multiple angles. From deMoivre's theorem, with $ninmathbbN$:
$$
beginalign
left( e^i theta right)^n &= e^i ntheta \
left( cos theta + i sin theta right)^n &= cos ntheta + i sin ntheta
endalign
$$
These intermediate formulas may help:
$$
beginalign
cos 2theta &= cos^2 theta + sin^2 theta \
sin 2theta &= 2 cos theta sin 2 theta \
endalign
$$
Reduce the denominator
$$
sin ^4(x)+cos ^4(x) = frac14 (cos (4 x)+3)
$$
The primitive is
$$
int frac1cos^4x + sin^4x , dx =
int frac1cos (4 x)+3 , dx =
left(4 sqrt2right)^-1arctan left(fractan (2 x)sqrt2right)
$$
Here is a look at the integrand:
answered Apr 17 '17 at 16:17
dantopa
6,02131640
add a comment |Â
up vote
0
down vote
Convert the exponential powers to multiple angles. From deMoivre's theorem, with $ninmathbbN$:
$$
beginalign
left( e^i theta right)^n &= e^i ntheta \
left( cos theta + i sin theta right)^n &= cos ntheta + i sin ntheta
endalign
$$
These intermediate formulas may help:
$$
beginalign
cos 2theta &= cos^2 theta + sin^2 theta \
sin 2theta &= 2 cos theta sin 2 theta \
endalign
$$
Reduce the denominator
$$
sin ^4(x)+cos ^4(x) = frac14 (cos (4 x)+3)
$$
The primitive is
$$
int frac1cos^4x + sin^4x , dx =
int frac1cos (4 x)+3 , dx =
left(4 sqrt2right)^-1arctan left(fractan (2 x)sqrt2right)
$$
Here is a look at the integrand:
answered Apr 17 '17 at 16:17
dantopa
6,02131640
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Convert the exponential powers to multiple angles. From deMoivre's theorem, with $ninmathbbN$:
$$
beginalign
left( e^i theta right)^n &= e^i ntheta \
left( cos theta + i sin theta right)^n &= cos ntheta + i sin ntheta
endalign
$$
These intermediate formulas may help:
$$
beginalign
cos 2theta &= cos^2 theta + sin^2 theta \
sin 2theta &= 2 cos theta sin 2 theta \
endalign
$$
Reduce the denominator
$$
sin ^4(x)+cos ^4(x) = frac14 (cos (4 x)+3)
$$
The primitive is
$$
int frac1cos^4x + sin^4x , dx =
int frac1cos (4 x)+3 , dx =
left(4 sqrt2right)^-1arctan left(fractan (2 x)sqrt2right)
$$
Here is a look at the integrand:
answered Apr 17 '17 at 16:17
dantopa
6,02131640
Convert the exponential powers to multiple angles. From deMoivre's theorem, with $ninmathbbN$:
$$
beginalign
left( e^i theta right)^n &= e^i ntheta \
left( cos theta + i sin theta right)^n &= cos ntheta + i sin ntheta
endalign
$$
These intermediate formulas may help:
$$
beginalign
cos 2theta &= cos^2 theta + sin^2 theta \
sin 2theta &= 2 cos theta sin 2 theta \
endalign
$$
Reduce the denominator
$$
sin ^4(x)+cos ^4(x) = frac14 (cos (4 x)+3)
$$
The primitive is
$$
int frac1cos^4x + sin^4x , dx =
int frac1cos (4 x)+3 , dx =
left(4 sqrt2right)^-1arctan left(fractan (2 x)sqrt2right)
$$
Here is a look at the integrand:
answered Apr 17 '17 at 16:17
dantopa
6,02131640
answered Apr 17 '17 at 16:17
dantopa
6,02131640
answered Apr 17 '17 at 16:17
dantopa
6,02131640
answered Apr 17 '17 at 16:17
dantopa
6,02131640
6,02131640
add a comment |Â
add a comment |Â
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Have you tried the Weierstrass substitution?
– Lucian
Jun 4 '14 at 19:44
Using the en.wikipedia.org/wiki/… will transform the integrand to the simpler $frac1a+bcos4t$.
– Yves Daoust
Jun 4 '14 at 19:48
8
Notice $$sin^4 x + cos^4x = 1 - 2sin^2xcos^2x = 1 - frac12sin^2(2x) = frac14(3+cos4x)$$ Introduce $t = tan2x$, we get $$intfracdxsin^4 x + cos^4x = int frac43 + frac1-t^21+t^2fracdt2(1+t^2) = intfracdt2+t^2 = frac1sqrt2tan^-1left(fractsqrt2right) + textconst.$$
– achille hui
Jun 4 '14 at 19:55