Question

How do you interpret $\int{x{{\sec }^{-1}}x}$ by integration by part method?

Answer 1

We are asked to integrate $x{{\sec }^{-1}}x$ by the part method. So, we will first consider one function as the first function and the other as the second function. Then we will apply $\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{\dfrac{dv}{dx}} \right)dx}$ . We will use the order ILATE rule. We will use the derivative of ${{\sec }^{-1}}x$ is given as $\dfrac{1}{x\sqrt{{{x}^{2}}-1}}.$

Complete step by step answer:
We are asked to evaluate the interval of $x{{\sec }^{-1}}x$ and we will use the integration by parts method. In this method, we will split our function into two parts and then we will choose which one is first and which one is the second function. To choose the first and second we will use the order as follows:
I – Inverse
L – log
A – Algebraic
T – Trigonometry
E – Exponent
So, in our function, $x{{\sec }^{-1}}x$ we will split it as x and ${{\sec }^{-1}}x.$ So, by the order given above, we get our first function is ${{\sec }^{-1}}x$ and the second function is x. As integration by parts is given as
$\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{\dfrac{dv}{dx}} \right)dx}$
Here ‘u’ represents the first function and ‘v’ represents the second function. So, we have $u={{\sec }^{-1}}x$ and v = x. Using these values, we get,
$\int{x{{\sec }^{-1}}xdx}={{\sec }^{-1}}x\int{x}dx-\int{\left( \dfrac{d{{\sec }^{-1}}x}{dx}\int{xdx} \right)dx}$
Now at $\int{xdx}=\dfrac{{{x}^{2}}}{2}$ and we know that $\dfrac{d\left( {{\sec }^{-1}}x \right)}{dx}=\dfrac{1}{x\sqrt{{{x}^{2}}-1}}.$
Using these values above, we get,
$\Rightarrow \dfrac{d\left( {{\sec }^{-1}}x \right)}{dx}={{\sec }^{-1}}x\times \dfrac{{{x}^{2}}}{2}-\int{\left[ \dfrac{{{x}^{2}}}{2}\times \dfrac{1}{x\sqrt{{{x}^{2}}-1}} \right]dx}$
On simplifying, we get,
$\Rightarrow \dfrac{d\left( {{\sec }^{-1}}x \right)}{dx}=\dfrac{{{x}^{2}}{{\sec }^{-1}}x}{2}-\dfrac{1}{2}\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}dx......\left( i \right)}$
Now, first, we will solve $\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}dx}$ separately. Now, let us assume ${{x}^{2}}-1$ as t. Then differentiating both the sides, we get,
$\Rightarrow 2xdx=dt$
So, using these in $\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}dx}$ we get
$\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}=\int{\dfrac{1}{2\sqrt{t}}dt}}$
Integrating we get,
$\Rightarrow \int{\dfrac{1}{2\sqrt{t}}dt}=\dfrac{1}{2}\int{{{t}^{\dfrac{1}{2}}}dt}$
As $\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}},$ so we get,
$\Rightarrow \int{\dfrac{1}{2\sqrt{t}}dt}=\dfrac{1}{2}\left[ \dfrac{{{t}^{\dfrac{-1}{2}+1}}}{\left( \dfrac{-1}{2}+1 \right)} \right]$
So, simplifying this, we get,
$\Rightarrow \int{\dfrac{1}{2\sqrt{t}}dt}=\dfrac{1}{2}\dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}$
$\Rightarrow \int{\dfrac{1}{2\sqrt{t}}dt}={{t}^{\dfrac{1}{2}}}$
Hence, we get,
$\Rightarrow \int{\dfrac{1}{2\sqrt{t}}dt}={{t}^{\dfrac{1}{2}}}$
Replacing the value back as $t={{x}^{2}}-1$ we get
$\int{\dfrac{x}{\sqrt{{{x}^{2}}-1}}dx}=\sqrt{{{x}^{2}}-1}$
Using this in (i), we get,
$=\dfrac{{{x}^{2}}{{\sec }^{-1}}x}{2}-\dfrac{1}{2}\left( \sqrt{{{x}^{2}}-1} \right)$
So, we get,
$\int{x{{\sec }^{-1}}xdx}=\dfrac{1}{2}{{x}^{2}}{{\sec }^{-1}}x-\dfrac{1}{2}\left( \sqrt{{{x}^{2}}-1} \right)+c$
where c is constant.

Note: Remember that in the case of indefinite integral, that is integral without limit, it is necessary to replace or substitute back the original value of the function back which we replaced earlier. While in the case of definite integral, no such thing is necessary as we change the function and also change the limit along with it. So, we were not required to change back the function. We simply apply the limit there with the change in function and the limit must also be changed accordingly.