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Question: How do you integrate \[{{\sec }^{-1}}x\]...

How do you integrate sec⁡−1x{{\sec }^{-1}}x

Explanation

Solution

The integration of the given function can be done by the integration of parts method. In this method we have to distribute the given variable into two parts and then we integrate them. This method is a little complicated if the student doesn't know the formula of the parts integration method.
We use the following formula for integration by parts:
∫[f(x)g(x)]dx=f(x)∫g(x)dx−∫[ddxf(x)∫g(x)dx]dx\int{\left[ f\left( x \right)g\left( x \right) \right]}dx=f\left( x \right)\int{g\left( x \right)dx-}\int{\left[ \dfrac{d}{dx}f\left( x \right)\int{g\left( x \right)dx} \right]}dx

Complete step by step solution:
The integral of the given function is in the form
I=∫sec⁡−1xdxI=\int{{{\sec }^{-1}}xdx}
To solve this integration it must have two functions so that we can apply integration by parts method. So the above integral can be written in the following form.
I=∫sec⁡−1x⋅1dx⋯⋯(1)I=\int{{{\sec }^{-1}}x\cdot 1dx}\cdots \cdots \left( 1 \right)
In equation (1) the first function is sec⁡−1x{{\sec }^{-1}}x and the second function is 1.
Now the formula of integration by parts is as follows:
∫[f(x)g(x)]dx=f(x)∫g(x)dx−∫[ddxf(x)∫g(x)dx]dx\int{\left[ f\left( x \right)g\left( x \right) \right]}dx=f\left( x \right)\int{g\left( x \right)dx-}\int{\left[ \dfrac{d}{dx}f\left( x \right)\int{g\left( x \right)dx} \right]}dx
Using the above formula equation (1) becomes
I=sec⁡−1x∫1dx−∫[ddxsec⁡−1x∫1dx]dxI={{\sec }^{-1}}x\int{1dx-\int{\left[ \dfrac{d}{dx}{{\sec }^{-1}}x\int{1dx} \right]}}dx
⇒I=xsec⁡−1x−∫[1xx2−1x]dx\Rightarrow I=x{{\sec }^{-1}}x-\int{\left[ \dfrac{1}{x\sqrt{{{x}^{2}}-1}}x \right]dx}
⇒I=xsec⁡−1x−∫1x2−1dx\Rightarrow I=x{{\sec }^{-1}}x-\int{\dfrac{1}{\sqrt{{{x}^{2}}-1}}dx}
We can also write the above expression in the following form
I=xsec⁡−1x−∫1x2−(1)2dx⋯⋯(2)I=x{{\sec }^{-1}}x-\int{\dfrac{1}{{{x}^{2}}-{{\left( 1 \right)}^{2}}}dx}\cdots \cdots \left( 2 \right)
We know that
∫1x2−a2dx=cosh⁡−1(xa)+c⋯⋯(3)\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx={{\cosh }^{-1}}\left( \dfrac{x}{a} \right)+c}\cdots \cdots \left( 3 \right)
From equation (2) and (3) we have
I=xsec⁡−1x−cosh⁡−1(x1)+cI=x{{\sec }^{-1}}x-{{\cosh }^{-1}}\left( \dfrac{x}{1} \right)+c

This is the final solution of the given problem.

Additional information:
The integration of sec⁡−1x{{\sec }^{-1}}x is an important integration function but there is no direct method to find out the integration of the sec⁡−1x{{\sec }^{-1}}x. So we use the integration by parts method. This method can be used to find the integration of complex functions also.

Note:
Students should have to pay attention while doing integration by parts.
Students should have carefully written the formula and substitute the values then carefully integrate the expression to get the final solution.