Integral of $int frac{cos^3x + sin^3x}{sqrt{cos 2x}}dx$ [closed]
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How do we integrate the following:
$$ intlimits_0^{pi/4} frac{cos^3x + sin^3x}{sqrt{cos 2x}}dx $$
I managed to find the existence of the integral but I can't find the way to solve this (calculus).
Extra : Sorry for my missing context. I will try to clear the problem. I read yours answer but I miss something because I have a problem with the upper integration limit $pi/4$. For me, there is a problem due to the fact that $cos (2*pi/4)= 0$.
Don't we need to proceed in the calculus of a limit.
calculus integration
asked Dec 27 '18 at 0:42
Sami KARIMSami KARIM
142
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closed as off-topic by Namaste, RRL, Zacky, Eevee Trainer, KReiser Dec 27 '18 at 8:45
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." – Namaste, RRL, Zacky, Eevee Trainer, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How do we integrate the following:
$$ intlimits_0^{pi/4} frac{cos^3x + sin^3x}{sqrt{cos 2x}}dx $$
I managed to find the existence of the integral but I can't find the way to solve this (calculus).
Extra : Sorry for my missing context. I will try to clear the problem. I read yours answer but I miss something because I have a problem with the upper integration limit $pi/4$. For me, there is a problem due to the fact that $cos (2*pi/4)= 0$.
Don't we need to proceed in the calculus of a limit.
calculus integration
asked Dec 27 '18 at 0:42
Sami KARIMSami KARIM
142
$endgroup$
closed as off-topic by Namaste, RRL, Zacky, Eevee Trainer, KReiser Dec 27 '18 at 8:45
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." – Namaste, RRL, Zacky, Eevee Trainer, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
4
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As Cleo would say, "$I = frac{1}{16}[-4+3pisqrt{2}+6sqrt{2}sinh^{-1}(1)]"$.
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– mathworker21
Dec 27 '18 at 0:47
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You should try $u=sinx$, and note $cos2x=cos^2x-sin^2x$ and see how far you can go.
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– herb steinberg
Dec 27 '18 at 0:49
2
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As Mathematica would say: $frac{1}{16} left(-4+3 sqrt{2} pi +6 sqrt{2} sinh ^{-1}(1)right)$.
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– David G. Stork
Dec 27 '18 at 0:55
3
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Note that $$int_0^{pi/4}frac{cos^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{cos(x)(1-sin^2(x))}{sqrt{1-2sin^2(x)}},dx=int_0^{1/sqrt2} frac{1-t^2}{sqrt{1-2t^2}},dt$$ $$int_0^{pi/4}frac{sin^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{sin(x)(1-cos^2(x))}{sqrt{2cos^2(x)-1}},dx=int_{1/sqrt2}^1 frac{1-t^2}{sqrt{2t^2-1}},dt$$
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– Mark Viola
Dec 27 '18 at 1:28
add a comment |
$begingroup$
How do we integrate the following:
$$ intlimits_0^{pi/4} frac{cos^3x + sin^3x}{sqrt{cos 2x}}dx $$
I managed to find the existence of the integral but I can't find the way to solve this (calculus).
Extra : Sorry for my missing context. I will try to clear the problem. I read yours answer but I miss something because I have a problem with the upper integration limit $pi/4$. For me, there is a problem due to the fact that $cos (2*pi/4)= 0$.
Don't we need to proceed in the calculus of a limit.
calculus integration
asked Dec 27 '18 at 0:42
Sami KARIMSami KARIM
142
$endgroup$
How do we integrate the following:
$$ intlimits_0^{pi/4} frac{cos^3x + sin^3x}{sqrt{cos 2x}}dx $$
I managed to find the existence of the integral but I can't find the way to solve this (calculus).
Extra : Sorry for my missing context. I will try to clear the problem. I read yours answer but I miss something because I have a problem with the upper integration limit $pi/4$. For me, there is a problem due to the fact that $cos (2*pi/4)= 0$.
Don't we need to proceed in the calculus of a limit.
calculus integration
calculus integration
asked Dec 27 '18 at 0:42
Sami KARIMSami KARIM
142
asked Dec 27 '18 at 0:42
Sami KARIMSami KARIM
142
edited Dec 27 '18 at 11:43
Sami KARIM
asked Dec 27 '18 at 0:42
Sami KARIMSami KARIM
142
asked Dec 27 '18 at 0:42
Sami KARIMSami KARIM
142
asked Dec 27 '18 at 0:42
Sami KARIMSami KARIM
142
142
closed as off-topic by Namaste, RRL, Zacky, Eevee Trainer, KReiser Dec 27 '18 at 8:45
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." – Namaste, RRL, Zacky, Eevee Trainer, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, RRL, Zacky, Eevee Trainer, KReiser Dec 27 '18 at 8:45
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." – Namaste, RRL, Zacky, Eevee Trainer, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
As Cleo would say, "$I = frac{1}{16}[-4+3pisqrt{2}+6sqrt{2}sinh^{-1}(1)]"$.
$endgroup$
– mathworker21
Dec 27 '18 at 0:47
$begingroup$
You should try $u=sinx$, and note $cos2x=cos^2x-sin^2x$ and see how far you can go.
$endgroup$
– herb steinberg
Dec 27 '18 at 0:49
2
$begingroup$
As Mathematica would say: $frac{1}{16} left(-4+3 sqrt{2} pi +6 sqrt{2} sinh ^{-1}(1)right)$.
$endgroup$
– David G. Stork
Dec 27 '18 at 0:55
3
$begingroup$
Note that $$int_0^{pi/4}frac{cos^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{cos(x)(1-sin^2(x))}{sqrt{1-2sin^2(x)}},dx=int_0^{1/sqrt2} frac{1-t^2}{sqrt{1-2t^2}},dt$$ $$int_0^{pi/4}frac{sin^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{sin(x)(1-cos^2(x))}{sqrt{2cos^2(x)-1}},dx=int_{1/sqrt2}^1 frac{1-t^2}{sqrt{2t^2-1}},dt$$
$endgroup$
– Mark Viola
Dec 27 '18 at 1:28
add a comment |
4
$begingroup$
As Cleo would say, "$I = frac{1}{16}[-4+3pisqrt{2}+6sqrt{2}sinh^{-1}(1)]"$.
$endgroup$
– mathworker21
Dec 27 '18 at 0:47
$begingroup$
You should try $u=sinx$, and note $cos2x=cos^2x-sin^2x$ and see how far you can go.
$endgroup$
– herb steinberg
Dec 27 '18 at 0:49
2
$begingroup$
As Mathematica would say: $frac{1}{16} left(-4+3 sqrt{2} pi +6 sqrt{2} sinh ^{-1}(1)right)$.
$endgroup$
– David G. Stork
Dec 27 '18 at 0:55
3
$begingroup$
Note that $$int_0^{pi/4}frac{cos^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{cos(x)(1-sin^2(x))}{sqrt{1-2sin^2(x)}},dx=int_0^{1/sqrt2} frac{1-t^2}{sqrt{1-2t^2}},dt$$ $$int_0^{pi/4}frac{sin^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{sin(x)(1-cos^2(x))}{sqrt{2cos^2(x)-1}},dx=int_{1/sqrt2}^1 frac{1-t^2}{sqrt{2t^2-1}},dt$$
$endgroup$
– Mark Viola
Dec 27 '18 at 1:28
4
4
$begingroup$
As Cleo would say, "$I = frac{1}{16}[-4+3pisqrt{2}+6sqrt{2}sinh^{-1}(1)]"$.
$endgroup$
– mathworker21
Dec 27 '18 at 0:47
$begingroup$
As Cleo would say, "$I = frac{1}{16}[-4+3pisqrt{2}+6sqrt{2}sinh^{-1}(1)]"$.
$endgroup$
– mathworker21
Dec 27 '18 at 0:47
$begingroup$
You should try $u=sinx$, and note $cos2x=cos^2x-sin^2x$ and see how far you can go.
$endgroup$
– herb steinberg
Dec 27 '18 at 0:49
$begingroup$
You should try $u=sinx$, and note $cos2x=cos^2x-sin^2x$ and see how far you can go.
$endgroup$
– herb steinberg
Dec 27 '18 at 0:49
2
2
$begingroup$
As Mathematica would say: $frac{1}{16} left(-4+3 sqrt{2} pi +6 sqrt{2} sinh ^{-1}(1)right)$.
$endgroup$
– David G. Stork
Dec 27 '18 at 0:55
$begingroup$
As Mathematica would say: $frac{1}{16} left(-4+3 sqrt{2} pi +6 sqrt{2} sinh ^{-1}(1)right)$.
$endgroup$
– David G. Stork
Dec 27 '18 at 0:55
3
3
$begingroup$
Note that $$int_0^{pi/4}frac{cos^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{cos(x)(1-sin^2(x))}{sqrt{1-2sin^2(x)}},dx=int_0^{1/sqrt2} frac{1-t^2}{sqrt{1-2t^2}},dt$$ $$int_0^{pi/4}frac{sin^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{sin(x)(1-cos^2(x))}{sqrt{2cos^2(x)-1}},dx=int_{1/sqrt2}^1 frac{1-t^2}{sqrt{2t^2-1}},dt$$
$endgroup$
– Mark Viola
Dec 27 '18 at 1:28
$begingroup$
Note that $$int_0^{pi/4}frac{cos^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{cos(x)(1-sin^2(x))}{sqrt{1-2sin^2(x)}},dx=int_0^{1/sqrt2} frac{1-t^2}{sqrt{1-2t^2}},dt$$ $$int_0^{pi/4}frac{sin^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{sin(x)(1-cos^2(x))}{sqrt{2cos^2(x)-1}},dx=int_{1/sqrt2}^1 frac{1-t^2}{sqrt{2t^2-1}},dt$$
$endgroup$
– Mark Viola
Dec 27 '18 at 1:28
add a comment |
1 Answer
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Following Steinberg's advice, substitute $u = sin x$. Note that $cos x = sqrt{1 - u^{2}}$ and $tan x = u/sqrt{1-u^{2}}$.
$$ I = int_{0}^{pi/4}frac{cos^{3}x + sin^{3}x}{sqrt{cos 2x}},mathrm{d}x = int_{0}^{1/sqrt{2}}left(frac{1-u^{2}}{sqrt{1-2u^{2}}} + frac{u^{3}}{sqrt{1-u^{2}}sqrt{1-2u^{2}}}right)mathrm{d}u$$
Substitute $x = 1 - 2u^{2}$.
$$ I = frac{1}{4sqrt{2}}left(int_{0}^{1}frac{1+x}{sqrt{x}sqrt{1-x}} + frac{1-x}{sqrt{x}sqrt{1+x}}right)mathrm{d}x $$
Using the beta function, the first term is evaluated as
$$begin{aligned} I_{1} &= frac{1}{4sqrt{2}}left(int_{0}^{1}x^{-1/2}(1-x)^{-1/2} + x^{1/2}(1-x)^{-1/2}right)mathrm{d}x \ &= frac{1}{4sqrt{2}}left(Gamma(1/2)Gamma(1/2) + Gamma(3/2)Gamma(1/2)right) = frac{3pi}{8sqrt{2}} end{aligned}$$
The second term is split into two integrals.
$$ I_{2} = frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} - frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x$$
These integrals are done by substituting $u = sqrt{x}$. The second integral requires the hyperbolic substitution $u = sinh y$ and can be done using hyperbolic identity, etc.
$$ frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} = frac{1}{2sqrt{2}}int_{0}^{1}frac{mathrm{d}u}{sqrt{1+u^{2}}} = frac{sinh^{-1}1}{2sqrt{2}}$$
$$begin{aligned} -frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x &= -frac{1}{2sqrt{2}}int_{0}^{1}frac{u^{2},mathrm{d}u}{sqrt{1+u^{2}}} \ &= -frac{1}{2sqrt{2}}int_{0}^{sinh^{-1}1}sinh^{2}y,mathrm{d}y \ &= -frac{1}{2sqrt{2}}left(frac{1}{sqrt{2}} - frac{sinh^{-1}1}{2}right) end{aligned} $$
The answer is thus
$$ boxed{I = frac{3pi}{8sqrt{2}} - frac{1}{4} + frac{3}{4sqrt{2}},sinh^{-1}1.} $$
answered Dec 27 '18 at 2:43
IninterrompueIninterrompue
67519
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Thanks for your answer i will try to study it on my side and say it if i meet some difficulties!
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– Sami KARIM
Dec 27 '18 at 11:01
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I edit the question can you clear my mind about the answer.
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– Sami KARIM
Dec 27 '18 at 11:44
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The divergence is weak because it is contained in a square root. The integral converges in a similar way as does $int_{0}^{1}1/sqrt{x},mathrm{d}x$.
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– Ininterrompue
Dec 27 '18 at 14:09
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Following Steinberg's advice, substitute $u = sin x$. Note that $cos x = sqrt{1 - u^{2}}$ and $tan x = u/sqrt{1-u^{2}}$.
$$ I = int_{0}^{pi/4}frac{cos^{3}x + sin^{3}x}{sqrt{cos 2x}},mathrm{d}x = int_{0}^{1/sqrt{2}}left(frac{1-u^{2}}{sqrt{1-2u^{2}}} + frac{u^{3}}{sqrt{1-u^{2}}sqrt{1-2u^{2}}}right)mathrm{d}u$$
Substitute $x = 1 - 2u^{2}$.
$$ I = frac{1}{4sqrt{2}}left(int_{0}^{1}frac{1+x}{sqrt{x}sqrt{1-x}} + frac{1-x}{sqrt{x}sqrt{1+x}}right)mathrm{d}x $$
Using the beta function, the first term is evaluated as
$$begin{aligned} I_{1} &= frac{1}{4sqrt{2}}left(int_{0}^{1}x^{-1/2}(1-x)^{-1/2} + x^{1/2}(1-x)^{-1/2}right)mathrm{d}x \ &= frac{1}{4sqrt{2}}left(Gamma(1/2)Gamma(1/2) + Gamma(3/2)Gamma(1/2)right) = frac{3pi}{8sqrt{2}} end{aligned}$$
The second term is split into two integrals.
$$ I_{2} = frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} - frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x$$
These integrals are done by substituting $u = sqrt{x}$. The second integral requires the hyperbolic substitution $u = sinh y$ and can be done using hyperbolic identity, etc.
$$ frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} = frac{1}{2sqrt{2}}int_{0}^{1}frac{mathrm{d}u}{sqrt{1+u^{2}}} = frac{sinh^{-1}1}{2sqrt{2}}$$
$$begin{aligned} -frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x &= -frac{1}{2sqrt{2}}int_{0}^{1}frac{u^{2},mathrm{d}u}{sqrt{1+u^{2}}} \ &= -frac{1}{2sqrt{2}}int_{0}^{sinh^{-1}1}sinh^{2}y,mathrm{d}y \ &= -frac{1}{2sqrt{2}}left(frac{1}{sqrt{2}} - frac{sinh^{-1}1}{2}right) end{aligned} $$
The answer is thus
$$ boxed{I = frac{3pi}{8sqrt{2}} - frac{1}{4} + frac{3}{4sqrt{2}},sinh^{-1}1.} $$
answered Dec 27 '18 at 2:43
IninterrompueIninterrompue
67519
$endgroup$
$begingroup$
Thanks for your answer i will try to study it on my side and say it if i meet some difficulties!
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:01
$begingroup$
I edit the question can you clear my mind about the answer.
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:44
$begingroup$
The divergence is weak because it is contained in a square root. The integral converges in a similar way as does $int_{0}^{1}1/sqrt{x},mathrm{d}x$.
$endgroup$
– Ininterrompue
Dec 27 '18 at 14:09
add a comment |
$begingroup$
Following Steinberg's advice, substitute $u = sin x$. Note that $cos x = sqrt{1 - u^{2}}$ and $tan x = u/sqrt{1-u^{2}}$.
$$ I = int_{0}^{pi/4}frac{cos^{3}x + sin^{3}x}{sqrt{cos 2x}},mathrm{d}x = int_{0}^{1/sqrt{2}}left(frac{1-u^{2}}{sqrt{1-2u^{2}}} + frac{u^{3}}{sqrt{1-u^{2}}sqrt{1-2u^{2}}}right)mathrm{d}u$$
Substitute $x = 1 - 2u^{2}$.
$$ I = frac{1}{4sqrt{2}}left(int_{0}^{1}frac{1+x}{sqrt{x}sqrt{1-x}} + frac{1-x}{sqrt{x}sqrt{1+x}}right)mathrm{d}x $$
Using the beta function, the first term is evaluated as
$$begin{aligned} I_{1} &= frac{1}{4sqrt{2}}left(int_{0}^{1}x^{-1/2}(1-x)^{-1/2} + x^{1/2}(1-x)^{-1/2}right)mathrm{d}x \ &= frac{1}{4sqrt{2}}left(Gamma(1/2)Gamma(1/2) + Gamma(3/2)Gamma(1/2)right) = frac{3pi}{8sqrt{2}} end{aligned}$$
The second term is split into two integrals.
$$ I_{2} = frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} - frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x$$
These integrals are done by substituting $u = sqrt{x}$. The second integral requires the hyperbolic substitution $u = sinh y$ and can be done using hyperbolic identity, etc.
$$ frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} = frac{1}{2sqrt{2}}int_{0}^{1}frac{mathrm{d}u}{sqrt{1+u^{2}}} = frac{sinh^{-1}1}{2sqrt{2}}$$
$$begin{aligned} -frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x &= -frac{1}{2sqrt{2}}int_{0}^{1}frac{u^{2},mathrm{d}u}{sqrt{1+u^{2}}} \ &= -frac{1}{2sqrt{2}}int_{0}^{sinh^{-1}1}sinh^{2}y,mathrm{d}y \ &= -frac{1}{2sqrt{2}}left(frac{1}{sqrt{2}} - frac{sinh^{-1}1}{2}right) end{aligned} $$
The answer is thus
$$ boxed{I = frac{3pi}{8sqrt{2}} - frac{1}{4} + frac{3}{4sqrt{2}},sinh^{-1}1.} $$
answered Dec 27 '18 at 2:43
IninterrompueIninterrompue
67519
$endgroup$
$begingroup$
Thanks for your answer i will try to study it on my side and say it if i meet some difficulties!
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:01
$begingroup$
I edit the question can you clear my mind about the answer.
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:44
$begingroup$
The divergence is weak because it is contained in a square root. The integral converges in a similar way as does $int_{0}^{1}1/sqrt{x},mathrm{d}x$.
$endgroup$
– Ininterrompue
Dec 27 '18 at 14:09
add a comment |
$begingroup$
Following Steinberg's advice, substitute $u = sin x$. Note that $cos x = sqrt{1 - u^{2}}$ and $tan x = u/sqrt{1-u^{2}}$.
$$ I = int_{0}^{pi/4}frac{cos^{3}x + sin^{3}x}{sqrt{cos 2x}},mathrm{d}x = int_{0}^{1/sqrt{2}}left(frac{1-u^{2}}{sqrt{1-2u^{2}}} + frac{u^{3}}{sqrt{1-u^{2}}sqrt{1-2u^{2}}}right)mathrm{d}u$$
Substitute $x = 1 - 2u^{2}$.
$$ I = frac{1}{4sqrt{2}}left(int_{0}^{1}frac{1+x}{sqrt{x}sqrt{1-x}} + frac{1-x}{sqrt{x}sqrt{1+x}}right)mathrm{d}x $$
Using the beta function, the first term is evaluated as
$$begin{aligned} I_{1} &= frac{1}{4sqrt{2}}left(int_{0}^{1}x^{-1/2}(1-x)^{-1/2} + x^{1/2}(1-x)^{-1/2}right)mathrm{d}x \ &= frac{1}{4sqrt{2}}left(Gamma(1/2)Gamma(1/2) + Gamma(3/2)Gamma(1/2)right) = frac{3pi}{8sqrt{2}} end{aligned}$$
The second term is split into two integrals.
$$ I_{2} = frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} - frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x$$
These integrals are done by substituting $u = sqrt{x}$. The second integral requires the hyperbolic substitution $u = sinh y$ and can be done using hyperbolic identity, etc.
$$ frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} = frac{1}{2sqrt{2}}int_{0}^{1}frac{mathrm{d}u}{sqrt{1+u^{2}}} = frac{sinh^{-1}1}{2sqrt{2}}$$
$$begin{aligned} -frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x &= -frac{1}{2sqrt{2}}int_{0}^{1}frac{u^{2},mathrm{d}u}{sqrt{1+u^{2}}} \ &= -frac{1}{2sqrt{2}}int_{0}^{sinh^{-1}1}sinh^{2}y,mathrm{d}y \ &= -frac{1}{2sqrt{2}}left(frac{1}{sqrt{2}} - frac{sinh^{-1}1}{2}right) end{aligned} $$
The answer is thus
$$ boxed{I = frac{3pi}{8sqrt{2}} - frac{1}{4} + frac{3}{4sqrt{2}},sinh^{-1}1.} $$
answered Dec 27 '18 at 2:43
IninterrompueIninterrompue
67519
$endgroup$
Following Steinberg's advice, substitute $u = sin x$. Note that $cos x = sqrt{1 - u^{2}}$ and $tan x = u/sqrt{1-u^{2}}$.
$$ I = int_{0}^{pi/4}frac{cos^{3}x + sin^{3}x}{sqrt{cos 2x}},mathrm{d}x = int_{0}^{1/sqrt{2}}left(frac{1-u^{2}}{sqrt{1-2u^{2}}} + frac{u^{3}}{sqrt{1-u^{2}}sqrt{1-2u^{2}}}right)mathrm{d}u$$
Substitute $x = 1 - 2u^{2}$.
$$ I = frac{1}{4sqrt{2}}left(int_{0}^{1}frac{1+x}{sqrt{x}sqrt{1-x}} + frac{1-x}{sqrt{x}sqrt{1+x}}right)mathrm{d}x $$
Using the beta function, the first term is evaluated as
$$begin{aligned} I_{1} &= frac{1}{4sqrt{2}}left(int_{0}^{1}x^{-1/2}(1-x)^{-1/2} + x^{1/2}(1-x)^{-1/2}right)mathrm{d}x \ &= frac{1}{4sqrt{2}}left(Gamma(1/2)Gamma(1/2) + Gamma(3/2)Gamma(1/2)right) = frac{3pi}{8sqrt{2}} end{aligned}$$
The second term is split into two integrals.
$$ I_{2} = frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} - frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x$$
These integrals are done by substituting $u = sqrt{x}$. The second integral requires the hyperbolic substitution $u = sinh y$ and can be done using hyperbolic identity, etc.
$$ frac{1}{4sqrt{2}}int_{0}^{1}frac{mathrm{d}x}{sqrt{x}sqrt{1+x}} = frac{1}{2sqrt{2}}int_{0}^{1}frac{mathrm{d}u}{sqrt{1+u^{2}}} = frac{sinh^{-1}1}{2sqrt{2}}$$
$$begin{aligned} -frac{1}{4sqrt{2}}int_{0}^{1}frac{sqrt{x}}{sqrt{1+x}},mathrm{d}x &= -frac{1}{2sqrt{2}}int_{0}^{1}frac{u^{2},mathrm{d}u}{sqrt{1+u^{2}}} \ &= -frac{1}{2sqrt{2}}int_{0}^{sinh^{-1}1}sinh^{2}y,mathrm{d}y \ &= -frac{1}{2sqrt{2}}left(frac{1}{sqrt{2}} - frac{sinh^{-1}1}{2}right) end{aligned} $$
The answer is thus
$$ boxed{I = frac{3pi}{8sqrt{2}} - frac{1}{4} + frac{3}{4sqrt{2}},sinh^{-1}1.} $$
answered Dec 27 '18 at 2:43
IninterrompueIninterrompue
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answered Dec 27 '18 at 2:43
IninterrompueIninterrompue
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answered Dec 27 '18 at 2:43
IninterrompueIninterrompue
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answered Dec 27 '18 at 2:43
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Thanks for your answer i will try to study it on my side and say it if i meet some difficulties!
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– Sami KARIM
Dec 27 '18 at 11:01
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I edit the question can you clear my mind about the answer.
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– Sami KARIM
Dec 27 '18 at 11:44
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The divergence is weak because it is contained in a square root. The integral converges in a similar way as does $int_{0}^{1}1/sqrt{x},mathrm{d}x$.
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– Ininterrompue
Dec 27 '18 at 14:09
add a comment |
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Thanks for your answer i will try to study it on my side and say it if i meet some difficulties!
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:01
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I edit the question can you clear my mind about the answer.
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:44
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The divergence is weak because it is contained in a square root. The integral converges in a similar way as does $int_{0}^{1}1/sqrt{x},mathrm{d}x$.
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– Ininterrompue
Dec 27 '18 at 14:09
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Thanks for your answer i will try to study it on my side and say it if i meet some difficulties!
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:01
$begingroup$
Thanks for your answer i will try to study it on my side and say it if i meet some difficulties!
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:01
$begingroup$
I edit the question can you clear my mind about the answer.
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:44
$begingroup$
I edit the question can you clear my mind about the answer.
$endgroup$
– Sami KARIM
Dec 27 '18 at 11:44
$begingroup$
The divergence is weak because it is contained in a square root. The integral converges in a similar way as does $int_{0}^{1}1/sqrt{x},mathrm{d}x$.
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– Ininterrompue
Dec 27 '18 at 14:09
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The divergence is weak because it is contained in a square root. The integral converges in a similar way as does $int_{0}^{1}1/sqrt{x},mathrm{d}x$.
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– Ininterrompue
Dec 27 '18 at 14:09
add a comment |
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As Cleo would say, "$I = frac{1}{16}[-4+3pisqrt{2}+6sqrt{2}sinh^{-1}(1)]"$.
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– mathworker21
Dec 27 '18 at 0:47
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You should try $u=sinx$, and note $cos2x=cos^2x-sin^2x$ and see how far you can go.
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– herb steinberg
Dec 27 '18 at 0:49
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As Mathematica would say: $frac{1}{16} left(-4+3 sqrt{2} pi +6 sqrt{2} sinh ^{-1}(1)right)$.
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– David G. Stork
Dec 27 '18 at 0:55
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Note that $$int_0^{pi/4}frac{cos^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{cos(x)(1-sin^2(x))}{sqrt{1-2sin^2(x)}},dx=int_0^{1/sqrt2} frac{1-t^2}{sqrt{1-2t^2}},dt$$ $$int_0^{pi/4}frac{sin^3(x)}{sqrt{cos(2x)}},dx=int_0^{pi/4}frac{sin(x)(1-cos^2(x))}{sqrt{2cos^2(x)-1}},dx=int_{1/sqrt2}^1 frac{1-t^2}{sqrt{2t^2-1}},dt$$
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– Mark Viola
Dec 27 '18 at 1:28