Evaluate $int frac{sqrt{64x^2-256}}{x},dx$
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$$int frac{sqrt{64x^2-256}}{x},dx$$
Image.
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
enter image description here
$newcommand{dd}{; mathrm{d}}int frac{sqrt{64x^2-256}}x dd x to
int frac{sqrt{64(x^2-4)}}x dd x to
int frac{8sqrt{x^2-4}}x dd x$
Use $x=asectheta$, $dd x=asectheta tantheta dd theta$.
$a=2$ $to$ $x=2sectheta$, $dd x=2sectheta tantheta dd theta$.
$=int frac{8sqrt{4sec^2theta-4}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8sqrt{4(sec^2theta-1)}}{2sectheta}(2secthetatantheta) dd theta $
$=int frac{8sqrt{4tan^2theta}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8(2tantheta)}{2sectheta}(2secthetatantheta) dd theta$
$=int 16tan^2theta dd theta to
16inttan^2theta dd theta to
underset{text{trig. formula}}{underbrace{16(theta+tantheta)+C}}$
$Rightarrow 16(tantheta-theta)+C = 16tantheta-16theta$
$x=2sectheta$, $sectheta= frac x2$
$boxed{16tanleft(frac{sqrt{x^2-4}}2right) -16sec^{-1}left(frac x2right)+C}$
calculus integration indefinite-integrals radicals
edited Jul 27 '16 at 8:50
Martin Sleziak
44.5k7115268
asked Jul 13 '16 at 4:03
cchang
11
add a comment |
up vote
0
down vote
favorite
$$int frac{sqrt{64x^2-256}}{x},dx$$
Image.
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
enter image description here
$newcommand{dd}{; mathrm{d}}int frac{sqrt{64x^2-256}}x dd x to
int frac{sqrt{64(x^2-4)}}x dd x to
int frac{8sqrt{x^2-4}}x dd x$
Use $x=asectheta$, $dd x=asectheta tantheta dd theta$.
$a=2$ $to$ $x=2sectheta$, $dd x=2sectheta tantheta dd theta$.
$=int frac{8sqrt{4sec^2theta-4}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8sqrt{4(sec^2theta-1)}}{2sectheta}(2secthetatantheta) dd theta $
$=int frac{8sqrt{4tan^2theta}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8(2tantheta)}{2sectheta}(2secthetatantheta) dd theta$
$=int 16tan^2theta dd theta to
16inttan^2theta dd theta to
underset{text{trig. formula}}{underbrace{16(theta+tantheta)+C}}$
$Rightarrow 16(tantheta-theta)+C = 16tantheta-16theta$
$x=2sectheta$, $sectheta= frac x2$
$boxed{16tanleft(frac{sqrt{x^2-4}}2right) -16sec^{-1}left(frac x2right)+C}$
calculus integration indefinite-integrals radicals
edited Jul 27 '16 at 8:50
Martin Sleziak
44.5k7115268
asked Jul 13 '16 at 4:03
cchang
11
Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
– Em.
Jul 13 '16 at 4:07
3
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
– Jack D'Aurizio
Jul 13 '16 at 4:09
I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
– Martin Sleziak
Jul 27 '16 at 8:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$int frac{sqrt{64x^2-256}}{x},dx$$
Image.
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
enter image description here
$newcommand{dd}{; mathrm{d}}int frac{sqrt{64x^2-256}}x dd x to
int frac{sqrt{64(x^2-4)}}x dd x to
int frac{8sqrt{x^2-4}}x dd x$
Use $x=asectheta$, $dd x=asectheta tantheta dd theta$.
$a=2$ $to$ $x=2sectheta$, $dd x=2sectheta tantheta dd theta$.
$=int frac{8sqrt{4sec^2theta-4}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8sqrt{4(sec^2theta-1)}}{2sectheta}(2secthetatantheta) dd theta $
$=int frac{8sqrt{4tan^2theta}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8(2tantheta)}{2sectheta}(2secthetatantheta) dd theta$
$=int 16tan^2theta dd theta to
16inttan^2theta dd theta to
underset{text{trig. formula}}{underbrace{16(theta+tantheta)+C}}$
$Rightarrow 16(tantheta-theta)+C = 16tantheta-16theta$
$x=2sectheta$, $sectheta= frac x2$
$boxed{16tanleft(frac{sqrt{x^2-4}}2right) -16sec^{-1}left(frac x2right)+C}$
calculus integration indefinite-integrals radicals
edited Jul 27 '16 at 8:50
Martin Sleziak
44.5k7115268
asked Jul 13 '16 at 4:03
cchang
11
$$int frac{sqrt{64x^2-256}}{x},dx$$
Image.
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
enter image description here
$newcommand{dd}{; mathrm{d}}int frac{sqrt{64x^2-256}}x dd x to
int frac{sqrt{64(x^2-4)}}x dd x to
int frac{8sqrt{x^2-4}}x dd x$
Use $x=asectheta$, $dd x=asectheta tantheta dd theta$.
$a=2$ $to$ $x=2sectheta$, $dd x=2sectheta tantheta dd theta$.
$=int frac{8sqrt{4sec^2theta-4}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8sqrt{4(sec^2theta-1)}}{2sectheta}(2secthetatantheta) dd theta $
$=int frac{8sqrt{4tan^2theta}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8(2tantheta)}{2sectheta}(2secthetatantheta) dd theta$
$=int 16tan^2theta dd theta to
16inttan^2theta dd theta to
underset{text{trig. formula}}{underbrace{16(theta+tantheta)+C}}$
$Rightarrow 16(tantheta-theta)+C = 16tantheta-16theta$
$x=2sectheta$, $sectheta= frac x2$
$boxed{16tanleft(frac{sqrt{x^2-4}}2right) -16sec^{-1}left(frac x2right)+C}$
calculus integration indefinite-integrals radicals
calculus integration indefinite-integrals radicals
edited Jul 27 '16 at 8:50
Martin Sleziak
44.5k7115268
asked Jul 13 '16 at 4:03
cchang
11
edited Jul 27 '16 at 8:50
Martin Sleziak
44.5k7115268
asked Jul 13 '16 at 4:03
cchang
11
edited Jul 27 '16 at 8:50
Martin Sleziak
44.5k7115268
edited Jul 27 '16 at 8:50
Martin Sleziak
44.5k7115268
edited Jul 27 '16 at 8:50
Martin Sleziak
44.5k7115268
44.5k7115268
asked Jul 13 '16 at 4:03
cchang
11
asked Jul 13 '16 at 4:03
cchang
11
asked Jul 13 '16 at 4:03
cchang
11
11
Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
– Em.
Jul 13 '16 at 4:07
3
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
– Jack D'Aurizio
Jul 13 '16 at 4:09
I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
– Martin Sleziak
Jul 27 '16 at 8:51
add a comment |
Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
– Em.
Jul 13 '16 at 4:07
3
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
– Jack D'Aurizio
Jul 13 '16 at 4:09
I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
– Martin Sleziak
Jul 27 '16 at 8:51
Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
– Em.
Jul 13 '16 at 4:07
Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
– Em.
Jul 13 '16 at 4:07
3
3
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
– Jack D'Aurizio
Jul 13 '16 at 4:09
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
– Jack D'Aurizio
Jul 13 '16 at 4:09
I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
– Martin Sleziak
Jul 27 '16 at 8:51
I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
– Martin Sleziak
Jul 27 '16 at 8:51
add a comment |
3 Answers
3
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oldest
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up vote
5
down vote
Since I am almost blind, I have a lot of problems reading the image.
Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$
answered Jul 13 '16 at 4:25
Claude Leibovici
117k1156131
Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
– cchang
Jul 13 '16 at 4:38
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
up vote
1
down vote
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$
$$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$
answered Jul 13 '16 at 6:05
lab bhattacharjee
221k15154271
@user376343,Thanks for your feedback
– lab bhattacharjee
2 days ago
add a comment |
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0
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$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
With the
sub$ds{ldots t equiv x - root{x^{2} - 4} imp
x = {t^{2} + 4 over 2t}}$:
begin{align}
&color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
8int{root{x^{2} - 4} over x},dd x =
8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
\[3mm] = &
32arctanpars{t over 2} + {128 over t} + 24t
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} +
32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
end{align}
answered Jul 13 '16 at 5:41
Felix Marin
66k7107139
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Since I am almost blind, I have a lot of problems reading the image.
Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$
answered Jul 13 '16 at 4:25
Claude Leibovici
117k1156131
Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
– cchang
Jul 13 '16 at 4:38
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
up vote
5
down vote
Since I am almost blind, I have a lot of problems reading the image.
Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$
answered Jul 13 '16 at 4:25
Claude Leibovici
117k1156131
Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
– cchang
Jul 13 '16 at 4:38
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
up vote
5
down vote
up vote
5
down vote
Since I am almost blind, I have a lot of problems reading the image.
Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$
answered Jul 13 '16 at 4:25
Claude Leibovici
117k1156131
Since I am almost blind, I have a lot of problems reading the image.
Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$
answered Jul 13 '16 at 4:25
Claude Leibovici
117k1156131
answered Jul 13 '16 at 4:25
Claude Leibovici
117k1156131
answered Jul 13 '16 at 4:25
Claude Leibovici
117k1156131
answered Jul 13 '16 at 4:25
Claude Leibovici
117k1156131
117k1156131
Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
– cchang
Jul 13 '16 at 4:38
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
– cchang
Jul 13 '16 at 4:38
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
– Claude Leibovici
Jul 13 '16 at 4:45
Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
– cchang
Jul 13 '16 at 4:38
Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
– cchang
Jul 13 '16 at 4:38
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
– Claude Leibovici
Jul 13 '16 at 4:45
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
up vote
1
down vote
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$
$$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$
answered Jul 13 '16 at 6:05
lab bhattacharjee
221k15154271
@user376343,Thanks for your feedback
– lab bhattacharjee
2 days ago
add a comment |
up vote
1
down vote
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$
$$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$
answered Jul 13 '16 at 6:05
lab bhattacharjee
221k15154271
@user376343,Thanks for your feedback
– lab bhattacharjee
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$
$$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$
answered Jul 13 '16 at 6:05
lab bhattacharjee
221k15154271
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$
$$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$
answered Jul 13 '16 at 6:05
lab bhattacharjee
221k15154271
edited 2 days ago
answered Jul 13 '16 at 6:05
lab bhattacharjee
221k15154271
answered Jul 13 '16 at 6:05
lab bhattacharjee
221k15154271
answered Jul 13 '16 at 6:05
lab bhattacharjee
221k15154271
221k15154271
@user376343,Thanks for your feedback
– lab bhattacharjee
2 days ago
add a comment |
@user376343,Thanks for your feedback
– lab bhattacharjee
2 days ago
@user376343,Thanks for your feedback
– lab bhattacharjee
2 days ago
@user376343,Thanks for your feedback
– lab bhattacharjee
2 days ago
add a comment |
up vote
0
down vote
$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
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newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
With the
sub$ds{ldots t equiv x - root{x^{2} - 4} imp
x = {t^{2} + 4 over 2t}}$:
begin{align}
&color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
8int{root{x^{2} - 4} over x},dd x =
8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
\[3mm] = &
32arctanpars{t over 2} + {128 over t} + 24t
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} +
32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
end{align}
answered Jul 13 '16 at 5:41
Felix Marin
66k7107139
add a comment |
up vote
0
down vote
$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
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newcommand{ul}[1]{underline{#1}}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
With the
sub$ds{ldots t equiv x - root{x^{2} - 4} imp
x = {t^{2} + 4 over 2t}}$:
begin{align}
&color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
8int{root{x^{2} - 4} over x},dd x =
8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
\[3mm] = &
32arctanpars{t over 2} + {128 over t} + 24t
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} +
32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
end{align}
answered Jul 13 '16 at 5:41
Felix Marin
66k7107139
add a comment |
up vote
0
down vote
up vote
0
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$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
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newcommand{ol}[1]{overline{#1}}
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newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
With the
sub$ds{ldots t equiv x - root{x^{2} - 4} imp
x = {t^{2} + 4 over 2t}}$:
begin{align}
&color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
8int{root{x^{2} - 4} over x},dd x =
8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
\[3mm] = &
32arctanpars{t over 2} + {128 over t} + 24t
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} +
32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
end{align}
answered Jul 13 '16 at 5:41
Felix Marin
66k7107139
$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
With the
sub$ds{ldots t equiv x - root{x^{2} - 4} imp
x = {t^{2} + 4 over 2t}}$:
begin{align}
&color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
8int{root{x^{2} - 4} over x},dd x =
8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
\[3mm] = &
32arctanpars{t over 2} + {128 over t} + 24t
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} +
32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
end{align}
answered Jul 13 '16 at 5:41
Felix Marin
66k7107139
answered Jul 13 '16 at 5:41
Felix Marin
66k7107139
answered Jul 13 '16 at 5:41
Felix Marin
66k7107139
answered Jul 13 '16 at 5:41
Felix Marin
66k7107139
66k7107139
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– Em.
Jul 13 '16 at 4:07
3
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
– Jack D'Aurizio
Jul 13 '16 at 4:09
I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
– Martin Sleziak
Jul 27 '16 at 8:51