Proving Integral of Arcsech x using Integration by Parts

[sooyong94]

Homework Statement

I was asked to prove the integral

$\int_{\frac{4}{5}}^{1} \textrm{arcsech}(x) =2\arctan 2-\frac{\pi}{2}-\frac{4}{5} \ln 2$

Homework Equations

Integration by parts

The Attempt at a Solution

Let $u=\textrm{arcsech} (x)$
$\textrm{sech u}=x$
$\textrm{cosh u}=\frac{1}{x}$
Differentiating implicitly,

$\textrm{sinh u} \frac{du}{dx}=\frac{-1}{x^{2}}$
$\frac{du}{dx}=\frac{-1}{x^{2}\textrm{sinh u}}$

Then I simplify it into
$\frac{-1}{x\sqrt{1-x^{2}}}$

Let $\frac{dv}{dx}=1$

$v=x$

Using integration by parts and evaluating the integral, I got
$\frac{\pi}{2}-\sin^{-1} \frac{4}{5} -\frac{4}{5} \ln 2$

Which is numerically correct. But how do I obtain $2\tan^{-1} 2$ as shown in the question above?

 

[Curious3141]

Homework Statement

I was asked to prove the integral

$\int_{\frac{4}{5}}^{1} \textrm{arcsech}(x) =2\arctan 2-\frac{\pi}{2}-\frac{4}{5} \ln 2$

Homework Equations

Integration by parts

The Attempt at a Solution

Let $u=\textrm{arcsech} (x)$
$\textrm{sech u}=x$
$\textrm{cosh u}=\frac{1}{x}$
Differentiating implicitly,

$\textrm{sinh u} \frac{du}{dx}=\frac{-1}{x^{2}}$
$\frac{du}{dx}=\frac{-1}{x^{2}\textrm{sinh u}}$

Then I simplify it into
$\frac{-1}{x\sqrt{1-x^{2}}}$

Let $\frac{dv}{dx}=1$

$v=x$

Using integration by parts and evaluating the integral, I got
$\frac{\pi}{2}-\sin^{-1} \frac{4}{5} -\frac{4}{5} \ln 2$

Which is numerically correct. But how do I obtain $2\tan^{-1} 2$ as shown in the question above?

Draw the standard 3-4-5 right triangle. Observe that $\frac{\pi}{2}-\sin^{-1} \frac{4}{5} = \tan^{-1}\frac{3}{4}$.

You now have to get $\tan^{-1}\frac{3}{4}$ into something with $\tan^{-1}2$ in it.

I found this a little tricky. The best solution I could find was to let:

$\tan^{-1}\frac{3}{4} = x$ so $\tan x = \frac{3}{4}$

Then let x = 2y so that $\tan 2y = \frac{3}{4}$

Solve for y in the form $y = \tan^{-1}z$, where z is something you have to find. Only one value is admissible. Express x as 2y = $2\tan^{-1}z.$

Now observe that $\frac{1 + \tan w}{1 - \tan w} = \tan(w + \frac{\pi}{4})$. Use that to find an alternative form for $\tan^{-1}z$, which will allow you to find x, in the form you need.

There might be a simpler way (indeed, it might start with an alternative solution of the integral), but I can't immediately find one.

 

[AlephZero]

To show the answers are the same, you have to show $2 \tan^{-1} 2 + \sin^{-1}\displaystyle\frac 4 5 = \pi$.

From a 3-4-5 triangle, $\sin^{-1}\displaystyle\frac 4 5 = \tan^{-1}\displaystyle\frac 4 3$.

From $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\displaystyle\frac{a+b}{1-ab}$,

$2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3$.

QED.

 

[sooyong94]

To show the answers are the same, you have to show $2 \tan^{-1} 2 + \sin^{-1}\displaystyle\frac 4 5 = \pi$.

From a 3-4-5 triangle, $\sin^{-1}\displaystyle\frac 4 5 = \tan^{-1}\displaystyle\frac 4 3$.

From $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\displaystyle\frac{a+b}{1-ab}$,

$2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3$.

QED.

I don't get it, but why
$2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3$.

 

[CAF123]

I think the equalities should be $$2 \tan^{-1} 2 = \tan^{-1} \left(-\frac{4}{3}\right) + \pi = \pi - \tan^{-1} \left(\frac{4}{3}\right)$$

Using Alephzero's formula with $a=b$ gives $$\tan(\tan^{-1} 2 + \tan^{-1} 2) = -\frac{4}{3} \Rightarrow 2\tan^{-1} 2 = \tan^{-1} \left(-\frac{4}{3}\right) + \pi$$

 

[SammyS]

Compare the graphs of $\ y=\text{arcsech}(x) \$ and $y=\text{sech}(x)\ .$

Integrate $y=\text{sech}(x)\$ to get an area related to that given by integrating $\ y=\text{arcsech}(x) \ .$

You will have to subtract the area of some rectangle.

 

[sooyong94]

Compare the graphs of $\ y=\text{arcsech}(x) \$ and $y=\text{sech}(x)\ .$

Integrate $y=\text{sech}(x)\$ to get an area related to that given by integrating $\ y=\text{arcsech}(x) \ .$

You will have to subtract the area of some rectangle.

I plotted two graphs and yet I can't figure it out... :(

 

[SammyS]

I plotted two graphs and yet I can't figure it out... :(

The definite integral you are evaluating represents the area below the y = arcsech(x) graph which is between x = 4/5 and x = 1 . Notice that arcsech(4/5) = ln(2) .

That is the same as the area below the y = sech(x) graph and above y = 4/5, for x ≥ 0. Right?