Mixing
Inverse trig function integration by parts.
$begingroup$
- $$int tan^{-1}{2y}dy$$
if I choose $u = tan^{-1}{2y}$ then $du = frac{2}{1 + (2y)^2} dy$
and
$dv = dy$ and $v = y$
But I have a more complicated du. What else can I do?
real-analysis integration
edited Jan 30 at 17:18
Larry
2,53031131
asked Jan 30 at 5:40
Jwan622Jwan622
2,34811632
$endgroup$
add a comment |
$begingroup$
- $$int tan^{-1}{2y}dy$$
if I choose $u = tan^{-1}{2y}$ then $du = frac{2}{1 + (2y)^2} dy$
and
$dv = dy$ and $v = y$
But I have a more complicated du. What else can I do?
real-analysis integration
edited Jan 30 at 17:18
Larry
2,53031131
asked Jan 30 at 5:40
Jwan622Jwan622
2,34811632
$endgroup$
$begingroup$
math.stackexchange.com/questions/768332/liate-how-does-it-work
$endgroup$
– lab bhattacharjee
Jan 30 at 5:50
add a comment |
$begingroup$
- $$int tan^{-1}{2y}dy$$
if I choose $u = tan^{-1}{2y}$ then $du = frac{2}{1 + (2y)^2} dy$
and
$dv = dy$ and $v = y$
But I have a more complicated du. What else can I do?
real-analysis integration
edited Jan 30 at 17:18
Larry
2,53031131
asked Jan 30 at 5:40
Jwan622Jwan622
2,34811632
$endgroup$
- $$int tan^{-1}{2y}dy$$
if I choose $u = tan^{-1}{2y}$ then $du = frac{2}{1 + (2y)^2} dy$
and
$dv = dy$ and $v = y$
But I have a more complicated du. What else can I do?
real-analysis integration
real-analysis integration
edited Jan 30 at 17:18
Larry
2,53031131
asked Jan 30 at 5:40
Jwan622Jwan622
2,34811632
edited Jan 30 at 17:18
Larry
2,53031131
asked Jan 30 at 5:40
Jwan622Jwan622
2,34811632
edited Jan 30 at 17:18
Larry
2,53031131
edited Jan 30 at 17:18
Larry
2,53031131
edited Jan 30 at 17:18
Larry
2,53031131
2,53031131
asked Jan 30 at 5:40
Jwan622Jwan622
2,34811632
asked Jan 30 at 5:40
Jwan622Jwan622
2,34811632
asked Jan 30 at 5:40
Jwan622Jwan622
2,34811632
2,34811632
$begingroup$
math.stackexchange.com/questions/768332/liate-how-does-it-work
$endgroup$
– lab bhattacharjee
Jan 30 at 5:50
add a comment |
$begingroup$
math.stackexchange.com/questions/768332/liate-how-does-it-work
$endgroup$
– lab bhattacharjee
Jan 30 at 5:50
$begingroup$
math.stackexchange.com/questions/768332/liate-how-does-it-work
$endgroup$
– lab bhattacharjee
Jan 30 at 5:50
$begingroup$
math.stackexchange.com/questions/768332/liate-how-does-it-work
$endgroup$
– lab bhattacharjee
Jan 30 at 5:50
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Another approach. Let $2y = tan(theta) rightarrow 2frac{dy}{dtheta} = sec^2(theta)$ and finally $dy = frac{1}{2}sec^2(theta)dtheta$.
Thus,
begin{align}
I =int arctan(2y):dy = int arctanleft(tan(theta)right) cdot frac{1}{2}sec^2(theta):dtheta = frac{1}{2}int theta sec^2(theta):dtheta
end{align}
Now, integrate by parts:
begin{align}
v'(theta) &= sec^2(theta) & u(theta) &= theta \
v(theta) &= tan(theta) & u'(theta) &= 1
end{align}
Thus:
begin{align}
I &= frac{1}{2}int theta sec^2(theta):dtheta = frac{1}{2}left[thetatan(theta) - int tan(theta) :dtheta right] \
&= frac{1}{2}left[thetatan(theta) - lnleft|sec(theta) right| right] + C
end{align}
Where $C$ is the constant of integration. Recall that $2y = tan(theta)$ and hence $theta = arctan(2y)$. For $sec(theta)$ we use the identity $sec(theta) = sqrt{tan^2(theta)+ 1} = sqrt{left(2yright)^2 + 1} = sqrt{4y^2 + 1}$. Hence we arrive at
begin{equation}
I = int arctan(2y):dy = frac{1}{2}left[2y arctan(2y) - lnleft| sqrt{4y^2 + 1} right|right] + C = y arctan(2y) - frac{1}{4}lnleft| 4y^2 + 1 right| + C
end{equation}
$endgroup$
add a comment |
$begingroup$
A neat identity.
Given some continuous inverse function $f^{-1}(x)$, we can find the antiderivative of this inverse function. Here's how.
Consider the integral
$$I=int f^{-1}(x)mathrm dx$$
Assuming that we know what $f(x)$ is, we may preform the substitution $x=f(t)$ meaning that $mathrm dx=f'(t)mathrm dt$. Hence we have
$$I=int f^{-1}left[f(t)right]f'(t)mathrm dt$$
$$I=int tf'(t)mathrm dt$$
We then integrate by parts:
$$mathrm dv=f'(t)mathrm dtRightarrow v=f(t)\
u=tRightarrow mathrm du=mathrm dt$$
Which gives
$$I=tf(t)-int f(t)mathrm dt$$
And assuming that $F'(x)=f(x)$,
$$I=tf(t)-F(t)+C$$
And since $x=f(t)$ we have that $t=f^{-1}(x)$ which gives
$$I=xf^{-1}(x)-Fleft[f^{-1}(x)right]+C$$
So, choosing $f^{-1}(x)=arctan ax$ for some constant $aneq 0$, we have that $f(x)=frac1atan x$. Hence
$$begin{align}
I&=xarctan ax-frac1aint tan t ,mathrm dt\
&=xarctan ax+frac1alnleft|cos tright|+C\
&=xarctan ax+frac1alnleft|cos[arctan ax]right|+C
end{align}$$
And from $$cos[arctan z]=frac1{sqrt{1+z^2}}$$
We have, after some simplification,
$$intarctan(ax)mathrm dx=xarctan(ax)-frac1{2a}ln(1+a^2x^2)+C$$
And your integral is given by the case $a=2$.
answered Jan 31 at 16:48
clathratusclathratus
5,0641439
$endgroup$
add a comment |
$begingroup$
$$intarctan2ydy=yarctan2y-intfrac{2y}{1+4y^2}dy=yarctan2y-frac{1}{4}ln(1+4y^2)+C.$$
answered Jan 30 at 5:45
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Integrating by parts, $int tan^{-1}(2y), dy=y, tan^{-1}(2y) -int frac {2y} {1+4y^{2}} , dy+c$. Put $1+4y^{2}=u$ to evaluate this.
answered Jan 30 at 5:47
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Another approach. Let $2y = tan(theta) rightarrow 2frac{dy}{dtheta} = sec^2(theta)$ and finally $dy = frac{1}{2}sec^2(theta)dtheta$.
Thus,
begin{align}
I =int arctan(2y):dy = int arctanleft(tan(theta)right) cdot frac{1}{2}sec^2(theta):dtheta = frac{1}{2}int theta sec^2(theta):dtheta
end{align}
Now, integrate by parts:
begin{align}
v'(theta) &= sec^2(theta) & u(theta) &= theta \
v(theta) &= tan(theta) & u'(theta) &= 1
end{align}
Thus:
begin{align}
I &= frac{1}{2}int theta sec^2(theta):dtheta = frac{1}{2}left[thetatan(theta) - int tan(theta) :dtheta right] \
&= frac{1}{2}left[thetatan(theta) - lnleft|sec(theta) right| right] + C
end{align}
Where $C$ is the constant of integration. Recall that $2y = tan(theta)$ and hence $theta = arctan(2y)$. For $sec(theta)$ we use the identity $sec(theta) = sqrt{tan^2(theta)+ 1} = sqrt{left(2yright)^2 + 1} = sqrt{4y^2 + 1}$. Hence we arrive at
begin{equation}
I = int arctan(2y):dy = frac{1}{2}left[2y arctan(2y) - lnleft| sqrt{4y^2 + 1} right|right] + C = y arctan(2y) - frac{1}{4}lnleft| 4y^2 + 1 right| + C
end{equation}
$endgroup$
add a comment |
$begingroup$
Another approach. Let $2y = tan(theta) rightarrow 2frac{dy}{dtheta} = sec^2(theta)$ and finally $dy = frac{1}{2}sec^2(theta)dtheta$.
Thus,
begin{align}
I =int arctan(2y):dy = int arctanleft(tan(theta)right) cdot frac{1}{2}sec^2(theta):dtheta = frac{1}{2}int theta sec^2(theta):dtheta
end{align}
Now, integrate by parts:
begin{align}
v'(theta) &= sec^2(theta) & u(theta) &= theta \
v(theta) &= tan(theta) & u'(theta) &= 1
end{align}
Thus:
begin{align}
I &= frac{1}{2}int theta sec^2(theta):dtheta = frac{1}{2}left[thetatan(theta) - int tan(theta) :dtheta right] \
&= frac{1}{2}left[thetatan(theta) - lnleft|sec(theta) right| right] + C
end{align}
Where $C$ is the constant of integration. Recall that $2y = tan(theta)$ and hence $theta = arctan(2y)$. For $sec(theta)$ we use the identity $sec(theta) = sqrt{tan^2(theta)+ 1} = sqrt{left(2yright)^2 + 1} = sqrt{4y^2 + 1}$. Hence we arrive at
begin{equation}
I = int arctan(2y):dy = frac{1}{2}left[2y arctan(2y) - lnleft| sqrt{4y^2 + 1} right|right] + C = y arctan(2y) - frac{1}{4}lnleft| 4y^2 + 1 right| + C
end{equation}
$endgroup$
add a comment |
$begingroup$
Another approach. Let $2y = tan(theta) rightarrow 2frac{dy}{dtheta} = sec^2(theta)$ and finally $dy = frac{1}{2}sec^2(theta)dtheta$.
Thus,
begin{align}
I =int arctan(2y):dy = int arctanleft(tan(theta)right) cdot frac{1}{2}sec^2(theta):dtheta = frac{1}{2}int theta sec^2(theta):dtheta
end{align}
Now, integrate by parts:
begin{align}
v'(theta) &= sec^2(theta) & u(theta) &= theta \
v(theta) &= tan(theta) & u'(theta) &= 1
end{align}
Thus:
begin{align}
I &= frac{1}{2}int theta sec^2(theta):dtheta = frac{1}{2}left[thetatan(theta) - int tan(theta) :dtheta right] \
&= frac{1}{2}left[thetatan(theta) - lnleft|sec(theta) right| right] + C
end{align}
Where $C$ is the constant of integration. Recall that $2y = tan(theta)$ and hence $theta = arctan(2y)$. For $sec(theta)$ we use the identity $sec(theta) = sqrt{tan^2(theta)+ 1} = sqrt{left(2yright)^2 + 1} = sqrt{4y^2 + 1}$. Hence we arrive at
begin{equation}
I = int arctan(2y):dy = frac{1}{2}left[2y arctan(2y) - lnleft| sqrt{4y^2 + 1} right|right] + C = y arctan(2y) - frac{1}{4}lnleft| 4y^2 + 1 right| + C
end{equation}
$endgroup$
Another approach. Let $2y = tan(theta) rightarrow 2frac{dy}{dtheta} = sec^2(theta)$ and finally $dy = frac{1}{2}sec^2(theta)dtheta$.
Thus,
begin{align}
I =int arctan(2y):dy = int arctanleft(tan(theta)right) cdot frac{1}{2}sec^2(theta):dtheta = frac{1}{2}int theta sec^2(theta):dtheta
end{align}
Now, integrate by parts:
begin{align}
v'(theta) &= sec^2(theta) & u(theta) &= theta \
v(theta) &= tan(theta) & u'(theta) &= 1
end{align}
Thus:
begin{align}
I &= frac{1}{2}int theta sec^2(theta):dtheta = frac{1}{2}left[thetatan(theta) - int tan(theta) :dtheta right] \
&= frac{1}{2}left[thetatan(theta) - lnleft|sec(theta) right| right] + C
end{align}
Where $C$ is the constant of integration. Recall that $2y = tan(theta)$ and hence $theta = arctan(2y)$. For $sec(theta)$ we use the identity $sec(theta) = sqrt{tan^2(theta)+ 1} = sqrt{left(2yright)^2 + 1} = sqrt{4y^2 + 1}$. Hence we arrive at
begin{equation}
I = int arctan(2y):dy = frac{1}{2}left[2y arctan(2y) - lnleft| sqrt{4y^2 + 1} right|right] + C = y arctan(2y) - frac{1}{4}lnleft| 4y^2 + 1 right| + C
end{equation}
answered Jan 30 at 6:36
user150203
add a comment |
add a comment |
$begingroup$
A neat identity.
Given some continuous inverse function $f^{-1}(x)$, we can find the antiderivative of this inverse function. Here's how.
Consider the integral
$$I=int f^{-1}(x)mathrm dx$$
Assuming that we know what $f(x)$ is, we may preform the substitution $x=f(t)$ meaning that $mathrm dx=f'(t)mathrm dt$. Hence we have
$$I=int f^{-1}left[f(t)right]f'(t)mathrm dt$$
$$I=int tf'(t)mathrm dt$$
We then integrate by parts:
$$mathrm dv=f'(t)mathrm dtRightarrow v=f(t)\
u=tRightarrow mathrm du=mathrm dt$$
Which gives
$$I=tf(t)-int f(t)mathrm dt$$
And assuming that $F'(x)=f(x)$,
$$I=tf(t)-F(t)+C$$
And since $x=f(t)$ we have that $t=f^{-1}(x)$ which gives
$$I=xf^{-1}(x)-Fleft[f^{-1}(x)right]+C$$
So, choosing $f^{-1}(x)=arctan ax$ for some constant $aneq 0$, we have that $f(x)=frac1atan x$. Hence
$$begin{align}
I&=xarctan ax-frac1aint tan t ,mathrm dt\
&=xarctan ax+frac1alnleft|cos tright|+C\
&=xarctan ax+frac1alnleft|cos[arctan ax]right|+C
end{align}$$
And from $$cos[arctan z]=frac1{sqrt{1+z^2}}$$
We have, after some simplification,
$$intarctan(ax)mathrm dx=xarctan(ax)-frac1{2a}ln(1+a^2x^2)+C$$
And your integral is given by the case $a=2$.
answered Jan 31 at 16:48
clathratusclathratus
5,0641439
$endgroup$
add a comment |
$begingroup$
A neat identity.
Given some continuous inverse function $f^{-1}(x)$, we can find the antiderivative of this inverse function. Here's how.
Consider the integral
$$I=int f^{-1}(x)mathrm dx$$
Assuming that we know what $f(x)$ is, we may preform the substitution $x=f(t)$ meaning that $mathrm dx=f'(t)mathrm dt$. Hence we have
$$I=int f^{-1}left[f(t)right]f'(t)mathrm dt$$
$$I=int tf'(t)mathrm dt$$
We then integrate by parts:
$$mathrm dv=f'(t)mathrm dtRightarrow v=f(t)\
u=tRightarrow mathrm du=mathrm dt$$
Which gives
$$I=tf(t)-int f(t)mathrm dt$$
And assuming that $F'(x)=f(x)$,
$$I=tf(t)-F(t)+C$$
And since $x=f(t)$ we have that $t=f^{-1}(x)$ which gives
$$I=xf^{-1}(x)-Fleft[f^{-1}(x)right]+C$$
So, choosing $f^{-1}(x)=arctan ax$ for some constant $aneq 0$, we have that $f(x)=frac1atan x$. Hence
$$begin{align}
I&=xarctan ax-frac1aint tan t ,mathrm dt\
&=xarctan ax+frac1alnleft|cos tright|+C\
&=xarctan ax+frac1alnleft|cos[arctan ax]right|+C
end{align}$$
And from $$cos[arctan z]=frac1{sqrt{1+z^2}}$$
We have, after some simplification,
$$intarctan(ax)mathrm dx=xarctan(ax)-frac1{2a}ln(1+a^2x^2)+C$$
And your integral is given by the case $a=2$.
answered Jan 31 at 16:48
clathratusclathratus
5,0641439
$endgroup$
add a comment |
$begingroup$
A neat identity.
Given some continuous inverse function $f^{-1}(x)$, we can find the antiderivative of this inverse function. Here's how.
Consider the integral
$$I=int f^{-1}(x)mathrm dx$$
Assuming that we know what $f(x)$ is, we may preform the substitution $x=f(t)$ meaning that $mathrm dx=f'(t)mathrm dt$. Hence we have
$$I=int f^{-1}left[f(t)right]f'(t)mathrm dt$$
$$I=int tf'(t)mathrm dt$$
We then integrate by parts:
$$mathrm dv=f'(t)mathrm dtRightarrow v=f(t)\
u=tRightarrow mathrm du=mathrm dt$$
Which gives
$$I=tf(t)-int f(t)mathrm dt$$
And assuming that $F'(x)=f(x)$,
$$I=tf(t)-F(t)+C$$
And since $x=f(t)$ we have that $t=f^{-1}(x)$ which gives
$$I=xf^{-1}(x)-Fleft[f^{-1}(x)right]+C$$
So, choosing $f^{-1}(x)=arctan ax$ for some constant $aneq 0$, we have that $f(x)=frac1atan x$. Hence
$$begin{align}
I&=xarctan ax-frac1aint tan t ,mathrm dt\
&=xarctan ax+frac1alnleft|cos tright|+C\
&=xarctan ax+frac1alnleft|cos[arctan ax]right|+C
end{align}$$
And from $$cos[arctan z]=frac1{sqrt{1+z^2}}$$
We have, after some simplification,
$$intarctan(ax)mathrm dx=xarctan(ax)-frac1{2a}ln(1+a^2x^2)+C$$
And your integral is given by the case $a=2$.
answered Jan 31 at 16:48
clathratusclathratus
5,0641439
$endgroup$
A neat identity.
Given some continuous inverse function $f^{-1}(x)$, we can find the antiderivative of this inverse function. Here's how.
Consider the integral
$$I=int f^{-1}(x)mathrm dx$$
Assuming that we know what $f(x)$ is, we may preform the substitution $x=f(t)$ meaning that $mathrm dx=f'(t)mathrm dt$. Hence we have
$$I=int f^{-1}left[f(t)right]f'(t)mathrm dt$$
$$I=int tf'(t)mathrm dt$$
We then integrate by parts:
$$mathrm dv=f'(t)mathrm dtRightarrow v=f(t)\
u=tRightarrow mathrm du=mathrm dt$$
Which gives
$$I=tf(t)-int f(t)mathrm dt$$
And assuming that $F'(x)=f(x)$,
$$I=tf(t)-F(t)+C$$
And since $x=f(t)$ we have that $t=f^{-1}(x)$ which gives
$$I=xf^{-1}(x)-Fleft[f^{-1}(x)right]+C$$
So, choosing $f^{-1}(x)=arctan ax$ for some constant $aneq 0$, we have that $f(x)=frac1atan x$. Hence
$$begin{align}
I&=xarctan ax-frac1aint tan t ,mathrm dt\
&=xarctan ax+frac1alnleft|cos tright|+C\
&=xarctan ax+frac1alnleft|cos[arctan ax]right|+C
end{align}$$
And from $$cos[arctan z]=frac1{sqrt{1+z^2}}$$
We have, after some simplification,
$$intarctan(ax)mathrm dx=xarctan(ax)-frac1{2a}ln(1+a^2x^2)+C$$
And your integral is given by the case $a=2$.
answered Jan 31 at 16:48
clathratusclathratus
5,0641439
answered Jan 31 at 16:48
clathratusclathratus
5,0641439
answered Jan 31 at 16:48
clathratusclathratus
5,0641439
answered Jan 31 at 16:48
clathratusclathratus
5,0641439
5,0641439
add a comment |
add a comment |
$begingroup$
$$intarctan2ydy=yarctan2y-intfrac{2y}{1+4y^2}dy=yarctan2y-frac{1}{4}ln(1+4y^2)+C.$$
answered Jan 30 at 5:45
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
$$intarctan2ydy=yarctan2y-intfrac{2y}{1+4y^2}dy=yarctan2y-frac{1}{4}ln(1+4y^2)+C.$$
answered Jan 30 at 5:45
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
$$intarctan2ydy=yarctan2y-intfrac{2y}{1+4y^2}dy=yarctan2y-frac{1}{4}ln(1+4y^2)+C.$$
answered Jan 30 at 5:45
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$$intarctan2ydy=yarctan2y-intfrac{2y}{1+4y^2}dy=yarctan2y-frac{1}{4}ln(1+4y^2)+C.$$
answered Jan 30 at 5:45
Michael RozenbergMichael Rozenberg
109k1896201
edited Jan 30 at 5:51
answered Jan 30 at 5:45
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 5:45
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 5:45
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
$begingroup$
Integrating by parts, $int tan^{-1}(2y), dy=y, tan^{-1}(2y) -int frac {2y} {1+4y^{2}} , dy+c$. Put $1+4y^{2}=u$ to evaluate this.
answered Jan 30 at 5:47
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
add a comment |
$begingroup$
Integrating by parts, $int tan^{-1}(2y), dy=y, tan^{-1}(2y) -int frac {2y} {1+4y^{2}} , dy+c$. Put $1+4y^{2}=u$ to evaluate this.
answered Jan 30 at 5:47
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
add a comment |
$begingroup$
Integrating by parts, $int tan^{-1}(2y), dy=y, tan^{-1}(2y) -int frac {2y} {1+4y^{2}} , dy+c$. Put $1+4y^{2}=u$ to evaluate this.
answered Jan 30 at 5:47
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
Integrating by parts, $int tan^{-1}(2y), dy=y, tan^{-1}(2y) -int frac {2y} {1+4y^{2}} , dy+c$. Put $1+4y^{2}=u$ to evaluate this.
answered Jan 30 at 5:47
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
edited Jan 30 at 8:13
answered Jan 30 at 5:47
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Jan 30 at 5:47
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Jan 30 at 5:47
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
71.9k53170
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add a comment |
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$begingroup$
math.stackexchange.com/questions/768332/liate-how-does-it-work
$endgroup$
– lab bhattacharjee
Jan 30 at 5:50