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Question: Find the integration of \[\int {{{\sec }^{ - 1}}x} \] by integration by parts method....

Find the integration of ∫sec⁡−1x\int {{{\sec }^{ - 1}}x} by integration by parts method.

Explanation

Solution

Hint : Integration by parts is a method to find integrals of products. To solve this integration it must have at least two functions; however this has only one function so the second function can be considered as “1”. After the consideration rule of by-pass can be successfully applied and a solution can be obtained.

Complete step-by-step answer :
The integral of secant inverse is an important integral function, but it has no direct method to find it. We shall solve the integration of secant inverse by using the integration by parts method.
The integral of secant inverse is of the form,
I=∫sec⁡−1xdxI = \int {{{\sec }^{ - 1}}x} dx
To solve this integration it must have at least two functions; however this has only one function so the second function can be considered as “1”. Now the integration becomes,
I=∫sec⁡−1x×1dxI = \int {{{\sec }^{ - 1}}x} \times 1dx
The first function is sec⁡−1x{\sec ^{ - 1}}x and the second function is 11 .
Using the formula for integration by parts we have,
∫[f(x)g(x)]dx=f(x)∫g(x)d(x)−∫[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)d\left( x \right) - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right] dx} } }
Using the formula above the equation becomes,

I=sec⁡−1x∫1dx−∫[ddxsec⁡−1x∫1dx]dx ⇒I=xsec⁡−1x−∫[1xx2−1x]dx ⇒I=xsec⁡−1x−∫1x2−1dx   I = {\sec ^{ - 1}}x\int {1dx - \int {\left[ {\dfrac{d}{{dx}}{{\sec }^{ - 1}}x\int {1dx} } \right] dx} } \\\ \Rightarrow I = x{\sec ^{ - 1}}x - \int {\left[ {\dfrac{1}{{x\sqrt {{x^2} - 1} }}x} \right] } dx \\\ \Rightarrow I = x{\sec ^{ - 1}}x - \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} \;

It can be written in the form
I=xsec⁡−1x−∫1x2−1dxI = x{\sec ^{ - 1}}x - \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}
Using the formula
∫1x2−a2dx=cosh⁡−1(xa)+c\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx = {{\cosh }^{ - 1}}\left( {\dfrac{x}{a}} \right) + c}
We have,

I=xsec⁡−1x−cosh⁡−1(x1)+c ⇒∫sec⁡−1xdx=xsec⁡−1x−cosh⁡−1x+c   I = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}\left( {\dfrac{x}{1}} \right) + c \\\ \Rightarrow \int {{{\sec }^{ - 1}}} xdx = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + c \;

Hence the integration is obtained by the by parts method. We can also use this integration of secant inverse as a formula.
So, the correct answer is “xsec⁡−1x−cosh⁡−1x+cx{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + c”.

Note : Identify “u” and “dv”. Priorities for choosing “u” are u=ln⁡xu = \ln x , u=xnu = {x^n} and u=eaxu = {e^{ax}} . “du” and “v” should be computed. The formula for the integration by parts should be used. To solve this integration it must have at least two functions. The second function can be considered as “1” if there is only one function.