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Question: If we are given an integral as \[\int{x{{\sec }^{-1}}xdx}=\dfrac{2}{k}\left[ {{x}^{2}}{{\sec }^{-1}}...

If we are given an integral as ∫xsec⁡−1xdx=2k[x2sec⁡−1x−x2−1]\int{x{{\sec }^{-1}}xdx}=\dfrac{2}{k}\left[ {{x}^{2}}{{\sec }^{-1}}x-\sqrt{{{x}^{2}}-1} \right] then find the value of k

Explanation

Solution

When we have given a product of two functions whether they are inverse functions or logarithmic functions or algebraic functions or trigonometric functions or exponential functions. Then we have to apply integration by parts formula by using the ILATE rule. So, in the integration by parts formula, there are two variables named as u and v, and variables u and v are taken by priority order in ILATE. If the inverse function comes first it should be taken as u and another corresponding function is taken as v

Complete step-by-step solution
let us take I as the given following integral in the left hand side
I=∫xsec⁡−1xdxI=\int{x{{\sec }^{-1}}xdx}
We know that the integration by parts formula is given by
∫uv=u∫v−∫du∫v\int{uv=u\int{v-\int{du\int{v}}}}
Let us consider u as x
Let us consider v as sec⁡−1x{{\sec }^{-1}}x
The u and v variables are taken using the ILATE rule
In this I stands for inverse function
L stands for logarithmic functions
A stands for algebraic functions
T stands for trigonometric functions
E stands for exponential function
in this problem there is one inverse function and other is algebraic function so, inverse function is taken as u and algebraic function is taken as v
Now apply integration by parts to the given integral then we will get
=sec⁡−1x(x22)−∫1xx2−1(x22)dx={{\sec }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{1}{x\sqrt{{{x}^{2}}-1}}}\left( \dfrac{{{x}^{2}}}{2} \right)dx
Let us assume that t=x2−1t={{x}^{2}}-1
Now apply derivative on both right hand side and left hand side we will get,
dt=2xdxdt=2xdx
Now replace the x2−1{{x}^{2}}-1by t and dx value by corresponding dt value we obtained. We will get as follows,
=sec⁡−1x(x22)−∫x2t(dt2x)={{\sec }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{x}{2\sqrt{t}}\left( \dfrac{dt}{2x} \right)}
Now cancelling the x in both numerator and denominator in integral part e will get,
=sec⁡−1x(x22)−14∫dtt={{\sec }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{4}\int{\dfrac{dt}{\sqrt{t}}}
=sec⁡−1x(x22)−14(2t)={{\sec }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{4}\left( 2\sqrt{t} \right)
Now again replace back t by x2−1{{x}^{2}}-1 we will get
=sec⁡−1x(x22)−14(2x2−1)={{\sec }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{4}\left( 2\sqrt{{{x}^{2}}-1} \right)
=12(x2sec⁡−1x−x2−1)=\dfrac{1}{2}\left( {{x}^{2}}{{\sec }^{-1}}x-\sqrt{{{x}^{2}}-1} \right)
So now compare the obtained equation and equation in right hand side given
⇒2k=12\Rightarrow \dfrac{2}{k}=\dfrac{1}{2}
⇒k=4\Rightarrow k=4

Note: Integration of sec⁡−1x{{\sec }^{-1}}x is give by 1xx2−1\dfrac{1}{x\sqrt{{{x}^{2}}-1}} and integration of 1t\dfrac{1}{\sqrt{t}} is given by 2t2\sqrt{t}. If the problem is given as an integral product of two functions named as algebraic function and trigonometric function. Then, the algebraic function is taken as u and trigonometric function is taken as v.in this way we can use ILATE rule