Evaluating $int cot x csc^2x ,mathrmdx$ with $u=cot x$
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$newcommanddmathrmd$
Evaluate the integral using the indicated substituion. $$int cot x csc^2x ,dx, qquad u= cot x .$$
Differentiating both sides of $u$, then making the substitution: $$
beginalign
u &=
phantom-cot x, \
d u &= -cot xcsc x ,dx, \
d x &= -fracd uu csc x.
endalign$$
$$int -fracucsc^2 x ,duucsc x = int -csc x ,du. $$
Apparently, this was not an adequate approach, because $x$ is still part of the integrand. What should be done instead?
calculus integration trigonometry indefinite-integrals
edited Sep 6 at 9:23
Nathanael Skrepek
1,5071515
asked Aug 30 at 17:39
Mauricio Mendes
968
 |Â
show 1 more comment
up vote
3
down vote
favorite
$newcommanddmathrmd$
Evaluate the integral using the indicated substituion. $$int cot x csc^2x ,dx, qquad u= cot x .$$
Differentiating both sides of $u$, then making the substitution: $$
beginalign
u &=
phantom-cot x, \
d u &= -cot xcsc x ,dx, \
d x &= -fracd uu csc x.
endalign$$
$$int -fracucsc^2 x ,duucsc x = int -csc x ,du. $$
Apparently, this was not an adequate approach, because $x$ is still part of the integrand. What should be done instead?
calculus integration trigonometry indefinite-integrals
edited Sep 6 at 9:23
Nathanael Skrepek
1,5071515
asked Aug 30 at 17:39
Mauricio Mendes
968
You have not completely turned the integrand to a function of $u$. Try do more.
– xbh
Aug 30 at 17:42
$csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
– xbh
Aug 30 at 17:44
2
Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
– xbh
Aug 30 at 17:51
I had confused it with another formula, but I got it now. Thank you, @xbh.
– Mauricio Mendes
Aug 30 at 17:58
2
$LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
– Mohammad Zuhair Khan
Aug 30 at 17:59
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$newcommanddmathrmd$
Evaluate the integral using the indicated substituion. $$int cot x csc^2x ,dx, qquad u= cot x .$$
Differentiating both sides of $u$, then making the substitution: $$
beginalign
u &=
phantom-cot x, \
d u &= -cot xcsc x ,dx, \
d x &= -fracd uu csc x.
endalign$$
$$int -fracucsc^2 x ,duucsc x = int -csc x ,du. $$
Apparently, this was not an adequate approach, because $x$ is still part of the integrand. What should be done instead?
calculus integration trigonometry indefinite-integrals
edited Sep 6 at 9:23
Nathanael Skrepek
1,5071515
asked Aug 30 at 17:39
Mauricio Mendes
968
$newcommanddmathrmd$
Evaluate the integral using the indicated substituion. $$int cot x csc^2x ,dx, qquad u= cot x .$$
Differentiating both sides of $u$, then making the substitution: $$
beginalign
u &=
phantom-cot x, \
d u &= -cot xcsc x ,dx, \
d x &= -fracd uu csc x.
endalign$$
$$int -fracucsc^2 x ,duucsc x = int -csc x ,du. $$
Apparently, this was not an adequate approach, because $x$ is still part of the integrand. What should be done instead?
calculus integration trigonometry indefinite-integrals
calculus integration trigonometry indefinite-integrals
edited Sep 6 at 9:23
Nathanael Skrepek
1,5071515
asked Aug 30 at 17:39
Mauricio Mendes
968
edited Sep 6 at 9:23
Nathanael Skrepek
1,5071515
asked Aug 30 at 17:39
Mauricio Mendes
968
edited Sep 6 at 9:23
Nathanael Skrepek
1,5071515
edited Sep 6 at 9:23
Nathanael Skrepek
1,5071515
edited Sep 6 at 9:23
Nathanael Skrepek
1,5071515
1,5071515
asked Aug 30 at 17:39
Mauricio Mendes
968
asked Aug 30 at 17:39
Mauricio Mendes
968
asked Aug 30 at 17:39
Mauricio Mendes
968
968
You have not completely turned the integrand to a function of $u$. Try do more.
– xbh
Aug 30 at 17:42
$csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
– xbh
Aug 30 at 17:44
2
Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
– xbh
Aug 30 at 17:51
I had confused it with another formula, but I got it now. Thank you, @xbh.
– Mauricio Mendes
Aug 30 at 17:58
2
$LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
– Mohammad Zuhair Khan
Aug 30 at 17:59
 |Â
show 1 more comment
You have not completely turned the integrand to a function of $u$. Try do more.
– xbh
Aug 30 at 17:42
$csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
– xbh
Aug 30 at 17:44
2
Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
– xbh
Aug 30 at 17:51
I had confused it with another formula, but I got it now. Thank you, @xbh.
– Mauricio Mendes
Aug 30 at 17:58
2
$LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
– Mohammad Zuhair Khan
Aug 30 at 17:59
You have not completely turned the integrand to a function of $u$. Try do more.
– xbh
Aug 30 at 17:42
You have not completely turned the integrand to a function of $u$. Try do more.
– xbh
Aug 30 at 17:42
$csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
– xbh
Aug 30 at 17:44
$csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
– xbh
Aug 30 at 17:44
2
2
Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
– xbh
Aug 30 at 17:51
Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
– xbh
Aug 30 at 17:51
I had confused it with another formula, but I got it now. Thank you, @xbh.
– Mauricio Mendes
Aug 30 at 17:58
I had confused it with another formula, but I got it now. Thank you, @xbh.
– Mauricio Mendes
Aug 30 at 17:58
2
2
$LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
– Mohammad Zuhair Khan
Aug 30 at 17:59
$LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
– Mohammad Zuhair Khan
Aug 30 at 17:59
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
$$
intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
$$
On the other hand, rewriting the integral as
$$
intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
$$
is much easier.
answered Aug 30 at 20:11
egreg
167k1281189
add a comment |Â
up vote
1
down vote
For alternative way:
$$int cot x csc^2 x dx$$
$$=int fraccos x dxsin^3 x$$
Now you can advance taking $sin x = z$ .
edited Aug 30 at 20:36
amWhy
190k27221433
answered Aug 30 at 18:02
Entrepreneur
339111
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
$$
intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
$$
On the other hand, rewriting the integral as
$$
intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
$$
is much easier.
answered Aug 30 at 20:11
egreg
167k1281189
add a comment |Â
up vote
3
down vote
accepted
You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
$$
intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
$$
On the other hand, rewriting the integral as
$$
intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
$$
is much easier.
answered Aug 30 at 20:11
egreg
167k1281189
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
$$
intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
$$
On the other hand, rewriting the integral as
$$
intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
$$
is much easier.
answered Aug 30 at 20:11
egreg
167k1281189
You have $du=-csc^2x,dx$, rather than your wrong differentiation. This implies the integral is
$$
intcot xcsc^2x,dx=int-u,du=-frac12u^2+c=-frac12cot^2x+c
$$
On the other hand, rewriting the integral as
$$
intfraccos xsin^3x,dx=int(sin x)^-3d(sin x)=-frac12frac1sin^2x+c
$$
is much easier.
answered Aug 30 at 20:11
egreg
167k1281189
answered Aug 30 at 20:11
egreg
167k1281189
answered Aug 30 at 20:11
egreg
167k1281189
answered Aug 30 at 20:11
egreg
167k1281189
167k1281189
add a comment |Â
add a comment |Â
up vote
1
down vote
For alternative way:
$$int cot x csc^2 x dx$$
$$=int fraccos x dxsin^3 x$$
Now you can advance taking $sin x = z$ .
edited Aug 30 at 20:36
amWhy
190k27221433
answered Aug 30 at 18:02
Entrepreneur
339111
add a comment |Â
up vote
1
down vote
For alternative way:
$$int cot x csc^2 x dx$$
$$=int fraccos x dxsin^3 x$$
Now you can advance taking $sin x = z$ .
edited Aug 30 at 20:36
amWhy
190k27221433
answered Aug 30 at 18:02
Entrepreneur
339111
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For alternative way:
$$int cot x csc^2 x dx$$
$$=int fraccos x dxsin^3 x$$
Now you can advance taking $sin x = z$ .
edited Aug 30 at 20:36
amWhy
190k27221433
answered Aug 30 at 18:02
Entrepreneur
339111
For alternative way:
$$int cot x csc^2 x dx$$
$$=int fraccos x dxsin^3 x$$
Now you can advance taking $sin x = z$ .
edited Aug 30 at 20:36
amWhy
190k27221433
answered Aug 30 at 18:02
Entrepreneur
339111
edited Aug 30 at 20:36
amWhy
190k27221433
edited Aug 30 at 20:36
amWhy
190k27221433
edited Aug 30 at 20:36
amWhy
190k27221433
190k27221433
answered Aug 30 at 18:02
Entrepreneur
339111
answered Aug 30 at 18:02
Entrepreneur
339111
answered Aug 30 at 18:02
Entrepreneur
339111
339111
add a comment |Â
add a comment |Â
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You have not completely turned the integrand to a function of $u$. Try do more.
– xbh
Aug 30 at 17:42
$csc^2(x) = 1 +cot^2(x) = 1 + u^2$.
– xbh
Aug 30 at 17:44
2
Also, $mathrm dcot(x) = -csc^2(x) mathrm dx$, not $-cot(x) csc (x)mathrm dx$.
– xbh
Aug 30 at 17:51
I had confused it with another formula, but I got it now. Thank you, @xbh.
– Mauricio Mendes
Aug 30 at 17:58
2
$LaTeX$ Tip: Try using cot x, csc x and mathrm dx to get $cot x$, $csc x$ and $mathrm dx$ respectively.
– Mohammad Zuhair Khan
Aug 30 at 17:59