Evaluate $int frac{sqrt{64x^2-256}}{x},dx$
$begingroup$
$$int frac{sqrt{64x^2-256}}{x},dx$$
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
44.7k9117272
asked Jul 13 '16 at 4:03
cchangcchang
11
$endgroup$
add a comment |
$begingroup$
$$int frac{sqrt{64x^2-256}}{x},dx$$
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
44.7k9117272
asked Jul 13 '16 at 4:03
cchangcchang
11
$endgroup$
$begingroup$
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.
$endgroup$
– Em.
Jul 13 '16 at 4:07
3
$begingroup$
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
$endgroup$
– Jack D'Aurizio
Jul 13 '16 at 4:09
$begingroup$
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.
$endgroup$
– Martin Sleziak
Jul 27 '16 at 8:51
add a comment |
$begingroup$
$$int frac{sqrt{64x^2-256}}{x},dx$$
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
44.7k9117272
asked Jul 13 '16 at 4:03
cchangcchang
11
$endgroup$
$$int frac{sqrt{64x^2-256}}{x},dx$$
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
44.7k9117272
asked Jul 13 '16 at 4:03
cchangcchang
11
edited Jul 27 '16 at 8:50
Martin Sleziak
44.7k9117272
asked Jul 13 '16 at 4:03
cchangcchang
11
edited Jul 27 '16 at 8:50
Martin Sleziak
44.7k9117272
edited Jul 27 '16 at 8:50
Martin Sleziak
44.7k9117272
edited Jul 27 '16 at 8:50
Martin Sleziak
44.7k9117272
44.7k9117272
asked Jul 13 '16 at 4:03
cchangcchang
11
asked Jul 13 '16 at 4:03
cchangcchang
11
asked Jul 13 '16 at 4:03
cchangcchang
11
11
$begingroup$
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.
$endgroup$
– Em.
Jul 13 '16 at 4:07
3
$begingroup$
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
$endgroup$
– Jack D'Aurizio
Jul 13 '16 at 4:09
$begingroup$
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.
$endgroup$
– Martin Sleziak
Jul 27 '16 at 8:51
add a comment |
$begingroup$
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.
$endgroup$
– Em.
Jul 13 '16 at 4:07
3
$begingroup$
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
$endgroup$
– Jack D'Aurizio
Jul 13 '16 at 4:09
$begingroup$
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.
$endgroup$
– Martin Sleziak
Jul 27 '16 at 8:51
$begingroup$
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.
$endgroup$
– Em.
Jul 13 '16 at 4:07
$begingroup$
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.
$endgroup$
– Em.
Jul 13 '16 at 4:07
3
3
$begingroup$
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
$endgroup$
– Jack D'Aurizio
Jul 13 '16 at 4:09
$begingroup$
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
$endgroup$
– Jack D'Aurizio
Jul 13 '16 at 4:09
$begingroup$
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.
$endgroup$
– Martin Sleziak
Jul 27 '16 at 8:51
$begingroup$
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.
$endgroup$
– Martin Sleziak
Jul 27 '16 at 8:51
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
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.
$endgroup$
– cchang
Jul 13 '16 at 4:38
$begingroup$
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
$endgroup$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
$begingroup$
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
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 bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
@user376343,Thanks for your feedback
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 17:07
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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 MarinFelix Marin
67.4k7107141
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
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.
$endgroup$
– cchang
Jul 13 '16 at 4:38
$begingroup$
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
$endgroup$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
$begingroup$
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 LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
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.
$endgroup$
– cchang
Jul 13 '16 at 4:38
$begingroup$
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
$endgroup$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
$begingroup$
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 LeiboviciClaude Leibovici
120k1157132
$endgroup$
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 LeiboviciClaude Leibovici
120k1157132
answered Jul 13 '16 at 4:25
Claude LeiboviciClaude Leibovici
120k1157132
answered Jul 13 '16 at 4:25
Claude LeiboviciClaude Leibovici
120k1157132
answered Jul 13 '16 at 4:25
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
$begingroup$
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.
$endgroup$
– cchang
Jul 13 '16 at 4:38
$begingroup$
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
$endgroup$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
$begingroup$
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.
$endgroup$
– cchang
Jul 13 '16 at 4:38
$begingroup$
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
$endgroup$
– Claude Leibovici
Jul 13 '16 at 4:45
$begingroup$
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.
$endgroup$
– cchang
Jul 13 '16 at 4:38
$begingroup$
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.
$endgroup$
– cchang
Jul 13 '16 at 4:38
$begingroup$
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
$endgroup$
– Claude Leibovici
Jul 13 '16 at 4:45
$begingroup$
If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
$endgroup$
– Claude Leibovici
Jul 13 '16 at 4:45
add a comment |
$begingroup$
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
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 bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
@user376343,Thanks for your feedback
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 17:07
add a comment |
$begingroup$
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
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 bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
@user376343,Thanks for your feedback
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 17:07
add a comment |
$begingroup$
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
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 bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$
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 bhattacharjeelab bhattacharjee
224k15156274
edited Nov 30 '18 at 17:06
answered Jul 13 '16 at 6:05
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jul 13 '16 at 6:05
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jul 13 '16 at 6:05
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
$begingroup$
@user376343,Thanks for your feedback
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 17:07
add a comment |
$begingroup$
@user376343,Thanks for your feedback
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 17:07
$begingroup$
@user376343,Thanks for your feedback
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 17:07
$begingroup$
@user376343,Thanks for your feedback
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 17:07
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
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newcommand{ic}{mathrm{i}}
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newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
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newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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 MarinFelix Marin
67.4k7107141
$endgroup$
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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 MarinFelix Marin
67.4k7107141
$endgroup$
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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 MarinFelix Marin
67.4k7107141
$endgroup$
$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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 MarinFelix Marin
67.4k7107141
answered Jul 13 '16 at 5:41
Felix MarinFelix Marin
67.4k7107141
answered Jul 13 '16 at 5:41
Felix MarinFelix Marin
67.4k7107141
answered Jul 13 '16 at 5:41
Felix MarinFelix Marin
67.4k7107141
67.4k7107141
add a comment |
add a comment |
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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.
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– Em.
Jul 13 '16 at 4:07
3
$begingroup$
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
$endgroup$
– Jack D'Aurizio
Jul 13 '16 at 4:09
$begingroup$
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.
$endgroup$
– Martin Sleziak
Jul 27 '16 at 8:51