
Find the integration of \[\int {{{\sec }^{ - 1}}x} \] by integration by parts method.
Answer
538.5k+ views
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,
\[I = \int {{{\sec }^{ - 1}}x} dx\]
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,
\[I = \int {{{\sec }^{ - 1}}x} \times 1dx\]
The first function is \[{\sec ^{ - 1}}x\] and the second function is \[1\] .
Using the formula for integration by parts we have,
\[\int {\left[ {f\left( x \right)g\left( x \right)} \right] dx = f\left( x \right)\int {g\left( x \right)d\left( x \right) - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right] dx} } } \]
Using the formula above the equation becomes,
\[
I = {\sec ^{ - 1}}x\int {1dx - \int {\left[ {\dfrac{d}{{dx}}{{\sec }^{ - 1}}x\int {1dx} } \right] dx} } \\
\Rightarrow I = x{\sec ^{ - 1}}x - \int {\left[ {\dfrac{1}{{x\sqrt {{x^2} - 1} }}x} \right] } dx \\
\Rightarrow I = x{\sec ^{ - 1}}x - \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} \;
\]
It can be written in the form
\[I = x{\sec ^{ - 1}}x - \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} \]
Using the formula
\[\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx = {{\cosh }^{ - 1}}\left( {\dfrac{x}{a}} \right) + c} \]
We have,
\[
I = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}\left( {\dfrac{x}{1}} \right) + c \\
\Rightarrow \int {{{\sec }^{ - 1}}} xdx = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + c \;
\]
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 “$x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + c$”.
Note: Identify “u” and “dv”. Priorities for choosing “u” are \[u = \ln x\] , \[u = {x^n}\] and \[u = {e^{ax}}\] . “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.
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,
\[I = \int {{{\sec }^{ - 1}}x} dx\]
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,
\[I = \int {{{\sec }^{ - 1}}x} \times 1dx\]
The first function is \[{\sec ^{ - 1}}x\] and the second function is \[1\] .
Using the formula for integration by parts we have,
\[\int {\left[ {f\left( x \right)g\left( x \right)} \right] dx = f\left( x \right)\int {g\left( x \right)d\left( x \right) - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right] dx} } } \]
Using the formula above the equation becomes,
\[
I = {\sec ^{ - 1}}x\int {1dx - \int {\left[ {\dfrac{d}{{dx}}{{\sec }^{ - 1}}x\int {1dx} } \right] dx} } \\
\Rightarrow I = x{\sec ^{ - 1}}x - \int {\left[ {\dfrac{1}{{x\sqrt {{x^2} - 1} }}x} \right] } dx \\
\Rightarrow I = x{\sec ^{ - 1}}x - \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} \;
\]
It can be written in the form
\[I = x{\sec ^{ - 1}}x - \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} \]
Using the formula
\[\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx = {{\cosh }^{ - 1}}\left( {\dfrac{x}{a}} \right) + c} \]
We have,
\[
I = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}\left( {\dfrac{x}{1}} \right) + c \\
\Rightarrow \int {{{\sec }^{ - 1}}} xdx = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + c \;
\]
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 “$x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + c$”.
Note: Identify “u” and “dv”. Priorities for choosing “u” are \[u = \ln x\] , \[u = {x^n}\] and \[u = {e^{ax}}\] . “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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

