Question

How do you integrate $\int {x{{\sec }^{ - 1}}xdx} $?

Answer 1

Hint: This question will be solved by integration by parts. In the integration by parts method if we integrate $f\left( x \right)g\left( x \right)$ we can write \[\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \int {f'\left( x \right)\left( {\int {g\left( x \right)dx} } \right)dx} \] where $f'\left( x \right) = \dfrac{{df\left( x \right)}}{{dx}}$. While choosing $f\left( x \right)$ and $g\left( x \right)$ choose in such a way that solving \[\int {f'\left( x \right)\left( {\int {g\left( x \right)dx} } \right)dx} \] would be easier. For example, while integrating $x{e^x}$ our $f\left( x \right)$ would be $x$ and $g\left( x \right)$ will be ${e^x}$ so that solving \[\int {f'\left( x \right)\left( {\int {g\left( x \right)dx} } \right)dx} \] is easier.

Complete step-by-step solution:
The integration by parts formula can be further written as integral of the product of any two functions = (First function × Integral of the second function) – Integral of [ (differentiation of the first function) × Integral of the second function]
The formula for integration by parts is
\[\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \int {f'\left( x \right)\left( {\int {g\left( x \right)dx} } \right)dx} \]
Where $f'\left( x \right) = \dfrac{{df\left( x \right)}}{{dx}}$.
In the given question we have to integrate $x{\sec ^{ - 1}}x$ by integration by parts method.
So, we can choose $f\left( x \right) = {\sec ^{ - 1}}x$ and $g\left( x \right) = x$.
Substituting the values in the formula for integration by parts, we get
\[ \Rightarrow \int {x{{\sec }^{ - 1}}xdx} = {\sec ^{ - 1}}x\int {xdx} - \int {\dfrac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right)\left( {\int {xdx} } \right)dx} \]
We know that $\dfrac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right) = \dfrac{1}{{x\sqrt {{x^2} - 1} }}$ and $\int {xdx} = \dfrac{{{x^2}}}{2}$. Substituting these values in the above equation,
\[ \Rightarrow \int {x{{\sec }^{ - 1}}xdx} = {\sec ^{ - 1}}x\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\dfrac{1}{{x\sqrt {{x^2} - 1} }} \times \dfrac{{{x^2}}}{2}dx} \]
Simplify the terms,
\[ \Rightarrow \int {x{{\sec }^{ - 1}}xdx} = \dfrac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \dfrac{1}{2}\int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} \]
Let $u = {x^2} - 1$. Then,
$ \Rightarrow du = 2xdx$
Divide both sides into both sides,
$ \Rightarrow \dfrac{1}{2}du = xdx$
Substitute the values,
\[ \Rightarrow \int {x{{\sec }^{ - 1}}xdx} = \dfrac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \dfrac{1}{2}\int {\dfrac{1}{{2\sqrt u }}du} \]
Simplify the terms,
\[ \Rightarrow \int {x{{\sec }^{ - 1}}xdx} = \dfrac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \dfrac{1}{4}\int {{u^{ - \dfrac{1}{2}}}du} \]
Integrate the terms,
\[ \Rightarrow \int {x{{\sec }^{ - 1}}xdx} = \dfrac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \dfrac{1}{4}\dfrac{{{u^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + C\]
Simplify the terms,
\[ \Rightarrow \int {x{{\sec }^{ - 1}}xdx} = \dfrac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \dfrac{{\sqrt u }}{2} + C\]
Substitute back the value of $u$,
\[ \Rightarrow \int {x{{\sec }^{ - 1}}xdx} = \dfrac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \dfrac{{\sqrt {{x^2} - 1} }}{2} + C\]

Hence, the integral of $x{\sec ^{ - 1}}x$ is \[\dfrac{{{x^2}{{\sec }^{ - 1}}x}}{2} - \dfrac{{\sqrt {{x^2} - 1} }}{2} + C\].

Note: Always remember the formula for integrating into the integration by parts method. Some people make mistakes while choosing f(x) and g(x) so carefully choose f(x) and g(x) such that integration would be easier to solve. Sometimes when we solve a problem, we find the function of L.H.S on the right-hand side, in that case, we should take our L.H.S as a variable like we did in the above question then it would be easier to solve. We just have to find the value of I.