Prove $int_0^inftyoperatorname{sech} x,dx=pi/2$, and deduce $int_0^1operatorname{sech}^{-1}x,dx$
$begingroup$
Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$
I can prove the first statement (see below), but I was unable to deduce the value of the second integral.
improper-integrals trigonometric-integrals
asked Jan 2 at 14:10
AbecAbec
145
$endgroup$
add a comment |
$begingroup$
Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$
I can prove the first statement (see below), but I was unable to deduce the value of the second integral.
improper-integrals trigonometric-integrals
asked Jan 2 at 14:10
AbecAbec
145
$endgroup$
$begingroup$
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
$endgroup$
– Matti P.
Jan 2 at 14:23
$begingroup$
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
$endgroup$
– Matti P.
Jan 2 at 14:30
$begingroup$
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
$endgroup$
– Abec
Jan 2 at 22:38
add a comment |
$begingroup$
Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$
I can prove the first statement (see below), but I was unable to deduce the value of the second integral.
improper-integrals trigonometric-integrals
asked Jan 2 at 14:10
AbecAbec
145
$endgroup$
Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$
I can prove the first statement (see below), but I was unable to deduce the value of the second integral.
improper-integrals trigonometric-integrals
improper-integrals trigonometric-integrals
asked Jan 2 at 14:10
AbecAbec
145
asked Jan 2 at 14:10
AbecAbec
145
edited Jan 4 at 1:41
Abec
asked Jan 2 at 14:10
AbecAbec
145
asked Jan 2 at 14:10
AbecAbec
145
asked Jan 2 at 14:10
AbecAbec
145
145
$begingroup$
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
$endgroup$
– Matti P.
Jan 2 at 14:23
$begingroup$
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
$endgroup$
– Matti P.
Jan 2 at 14:30
$begingroup$
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
$endgroup$
– Abec
Jan 2 at 22:38
add a comment |
$begingroup$
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
$endgroup$
– Matti P.
Jan 2 at 14:23
$begingroup$
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
$endgroup$
– Matti P.
Jan 2 at 14:30
$begingroup$
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
$endgroup$
– Abec
Jan 2 at 22:38
$begingroup$
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
$endgroup$
– Matti P.
Jan 2 at 14:23
$begingroup$
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
$endgroup$
– Matti P.
Jan 2 at 14:23
$begingroup$
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
$endgroup$
– Matti P.
Jan 2 at 14:30
$begingroup$
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
$endgroup$
– Matti P.
Jan 2 at 14:30
$begingroup$
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
$endgroup$
– Abec
Jan 2 at 22:38
$begingroup$
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
$endgroup$
– Abec
Jan 2 at 22:38
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered Jan 2 at 22:42
Jacky ChongJacky Chong
19.3k21129
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add a comment |
$begingroup$
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered Jan 2 at 22:42
Jacky ChongJacky Chong
19.3k21129
$endgroup$
add a comment |
$begingroup$
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered Jan 2 at 22:42
Jacky ChongJacky Chong
19.3k21129
$endgroup$
add a comment |
$begingroup$
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered Jan 2 at 22:42
Jacky ChongJacky Chong
19.3k21129
$endgroup$
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered Jan 2 at 22:42
Jacky ChongJacky Chong
19.3k21129
edited Jan 2 at 23:13
answered Jan 2 at 22:42
Jacky ChongJacky Chong
19.3k21129
answered Jan 2 at 22:42
Jacky ChongJacky Chong
19.3k21129
answered Jan 2 at 22:42
Jacky ChongJacky Chong
19.3k21129
19.3k21129
add a comment |
add a comment |
$begingroup$
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
$endgroup$
add a comment |
$begingroup$
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
$endgroup$
add a comment |
$begingroup$
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
$endgroup$
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
answered Jan 3 at 8:37
user150203
add a comment |
add a comment |
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$begingroup$
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
$endgroup$
– Matti P.
Jan 2 at 14:23
$begingroup$
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
$endgroup$
– Matti P.
Jan 2 at 14:30
$begingroup$
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
$endgroup$
– Abec
Jan 2 at 22:38