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How do you integrate $\int{x \text{ arcsec} x}$ using integration by parts?

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Last updated date: 03rd Mar 2024
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Hint: Take $\text{arcsec} x$ as the first function and ‘x’ as the second function for the integration by parts rule. Put the the integration of \[\int{x}dx\] as \[\dfrac{{{x}^{2}}}{2}\] and the derivative of \[\dfrac{d}{dx}\left( \text{arcsec} x \right)\] as \[\dfrac{1}{x\sqrt{{{x}^{2}}-1}}\]. Solve the integration \[\int{\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)}dx\] separately by taking $u={{x}^{2}}-1$ and put the value of the integration later to obtain the required solution.

Complete step-by-step solution:
Integration by parts: If we have two functions let say ‘u’ and ‘v’, then the integration $\int{u.v}$ can be carried out using integration by parts as $\int{u.v}=u\int{vdv-\int{\left( \dfrac{du}{dx}\int{v}dv \right)}}dv$.
Now, considering our equation, $\int{x\text{arcsec} x}$
Here $u=\text{arcsec} x$ and $v=x$
Applying by parts, we get
\[\begin{align}
  & \int{\text{arcsec} x}\cdot x \\
 & \Rightarrow \text{arcsec} x\int{x}dx-\int{\left( \dfrac{d}{dx}\left( \text{arcsec} x \right)\int{x}dx \right)dx} \\
\end{align}\]
As we know, the integration of \[\int{x}dx=\dfrac{{{x}^{2}}}{2}\] and the derivative of \[\dfrac{d}{dx}\left( \text{arcsec} x \right)=\dfrac{1}{x\sqrt{{{x}^{2}}-1}}\]
So substituting these values, we get
\[\begin{align}
  & \Rightarrow \text{arcsec} x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\left( \dfrac{1}{x\sqrt{{{x}^{2}}-1}} \right)\left( \dfrac{{{x}^{2}}}{2} \right)}dx \\
 & \Rightarrow \text{arcsec} x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\int{\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)}dx \\
\end{align}\]
For solving the integration \[\int{\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)}dx\]
Let $u={{x}^{2}}-1$
$\begin{align}
  & \Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( {{x}^{2}}-1 \right) \\
 & \Rightarrow du=2xdx \\
 & \Rightarrow xdx=\dfrac{1}{2}du \\
\end{align}$
Replacing the value of ${{x}^{2}}-1$ as ‘u’ and $xdx$ as $\dfrac{1}{2}du$in the integration \[\int{\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)}dx\], we get
\[\begin{align}
  & \Rightarrow \int{\dfrac{\dfrac{1}{2}du}{\sqrt{u}}} \\
 & \Rightarrow \dfrac{1}{2}\int{{{u}^{-\dfrac{1}{2}}}}du \\
 & \Rightarrow \dfrac{1}{2}\dfrac{{{u}^{-\dfrac{1}{2}+1}}}{^{-\dfrac{1}{2}+1}} \\
 & \Rightarrow \dfrac{1}{2}\dfrac{{{u}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \\
 & \Rightarrow \dfrac{1}{2}\times \dfrac{2}{1}{{u}^{\dfrac{1}{2}}} \\
 & \Rightarrow \sqrt{u} \\
\end{align}\]
Substituting the value of ‘u’, we get
\[\int{\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)}dx=\sqrt{{{x}^{2}}-1}\]
Coming back to our solution, putting \[\int{\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)}dx=\sqrt{{{x}^{2}}-1}\], we get
\[\Rightarrow \dfrac{1}{2}{{x}^{2}}\text{arcsec} x-\dfrac{1}{2}\sqrt{{{x}^{2}}-1}+c\]
This is the required solution of the given question.

Note: For indefinite integral one constant ‘c’ should be added to the final result. In integration by parts the first function ‘u’ and the second function ‘v’ is decided according to the order of ILATE rule which stands for Inverse Logarithmic Algebraic Trigonometric Exponential. So for our integration $\text{arcsec} x$ is considered as the first function as it is an inverse function. Similarly ‘x’ is considered as the second function as it is an algebraic function.



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