Question

.............................Evaluate...........................​

Answer 1

We have to find,

$\quad$

$\displaystyle\longrightarrow\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx=\,?}$

$\quad$

Multiplying and dividing denominator by a,

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\int\dfrac {1}{\left (\dfrac {a|x|\sqrt{x^2-a^2}}{a}\right)}dx}\end {aligned}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\int\dfrac {1}{a|x|\sqrt{\dfrac {x^2-a^2}{a^2}}}dx}\end {aligned}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\int\dfrac {1}{a|x|\sqrt{\left (\dfrac {x}{a}\right)^2-1}}dx\quad\quad\dots (1)}\end {aligned}$

$\quad$

Let,

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$\displaystyle\longrightarrow\sf{u=\dfrac {x}{a}}$

$\quad$

Since a is constant,

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$\displaystyle\longrightarrow\sf{\dfrac {du}{dx}=\dfrac {1}{a}}$

$\quad$

$\displaystyle\longrightarrow\sf{dx=a\ du}$

$\quad$

Then (1) becomes,

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\int\dfrac {a}{a|au|\sqrt{u^2-1}}du}\end {aligned}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\int\dfrac {1}{a|u|\sqrt{u^2-1}}du}\end {aligned}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\dfrac {1}{a}\int\dfrac {1}{|u|\sqrt{u^2-1}}du}\end {aligned}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\dfrac {1}{a}\sec^{-1}(u)+C}\end {aligned}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\dfrac {1}{a}\sec^{-1}\left (\dfrac{x}{a}\right)+C}\end {aligned}$

$\quad$

Since $\displaystyle\sf{\sec^{-1}(t)=\dfrac {\pi}{2}-\csc^{-1}(t),}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\dfrac {1}{a}\left (\dfrac {\pi}{2}-\csc^{-1}\left (\dfrac{x}{a}\right)\right)+C}\end {aligned}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{\dfrac {\pi}{2a}-\dfrac {1}{a}\csc^{-1}\left (\dfrac{x}{a}\right)+C}\end {aligned}$

$\quad$

Taking $\displaystyle\sf{\dfrac {\pi}{2a}+C=K,}$

$\quad$

$\begin {aligned}\displaystyle\longrightarrow\ \ &\sf{\int\dfrac {1}{|x|\sqrt{x^2-a^2}}dx}\\\\=\ \ &\sf{-\dfrac {1}{a}\csc^{-1}\left (\dfrac{x}{a}\right)+K}\end {aligned}$