
How do you integrate $\int {x\,{{\sec }^2}\,x} $ by integration by parts method?
Answer
546.9k+ views
Hint: We have a question where we are supposed to integrate the function with the help of integration by parts method. It is like the product rule for integration. In fact, it is also derived from the product rule for differentiation.
The rule is:
$\int {udv = uv - \int {vdu} } $
Complete step-by-step solution:
We have a function $\int {x\,{{\sec }^2}\,x} $ , and to find the integration by parts, certain steps are to be followed,
This is a classic case of integration by parts, which takes the form
\[\smallint udv = uv - \smallint vdu\]
For the given integral \[\smallint x\,se{c^2}(x)dx\] , we want to choose a value of $u$ that gets simpler when we differentiate it and a value of \[dv\] that is easily integrated. So, we will substitute the value of the given equation in the formula that we have,
So, let:
$u = x, \Rightarrow du = dx$
$v = \tan x, \Rightarrow dv = {\sec ^2}x$
We then have,
\[ \Rightarrow \smallint x\,{\sec ^2}(x)dx = uv - \smallint vdu\]
Putting the values in the formula,
$ \Rightarrow x\tan (x) - \smallint \tan (x)dx$
You may be aware of the integral of \[tan(x)\;\] . If not, it's easy to find:
\[ \Rightarrow xtan(x) - \int {\dfrac{{\sin (x)}}{{\cos (x)}}} dx\]
Let \[t = cos(x)\] , implying that \[dt = - sin(x)dx\]
\[ \Rightarrow \smallint x{\sec ^2}(x)dx = x\tan (x) + \smallint \dfrac{{ - \sin (x)}}{{\cos (x)}}dx\]
Solving the equation, until it is simplified,
\[ \Rightarrow \smallint x{\sec ^2}(x)dx = x\tan (x) + \smallint \dfrac{1}{t}dt\]
This is a common integral:
\[ \Rightarrow \smallint x{\sec ^2}(x)dx = x\tan (x) + \ln \left| t \right| + C\]
Working back from \[t = cos(x)\]:
\[\smallint x{\sec ^2}(x)dx = x\tan (x) + \ln |\cos (x)| + C\;\]
Therefore, for an equation $\int {x\,{{\sec }^2}\,x} $ the integration by parts will make the solution come up to $x\tan (x) + \ln |\cos x| + C$
Note: Integration is a way of adding the slices to add the whole. It is the calculation of an integral. Integration by parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. Integration by parts is basically used for functions that can also be constructed as products of some other function and a third function’s derivative.
The rule is:
$\int {udv = uv - \int {vdu} } $
Complete step-by-step solution:
We have a function $\int {x\,{{\sec }^2}\,x} $ , and to find the integration by parts, certain steps are to be followed,
This is a classic case of integration by parts, which takes the form
\[\smallint udv = uv - \smallint vdu\]
For the given integral \[\smallint x\,se{c^2}(x)dx\] , we want to choose a value of $u$ that gets simpler when we differentiate it and a value of \[dv\] that is easily integrated. So, we will substitute the value of the given equation in the formula that we have,
So, let:
$u = x, \Rightarrow du = dx$
$v = \tan x, \Rightarrow dv = {\sec ^2}x$
We then have,
\[ \Rightarrow \smallint x\,{\sec ^2}(x)dx = uv - \smallint vdu\]
Putting the values in the formula,
$ \Rightarrow x\tan (x) - \smallint \tan (x)dx$
You may be aware of the integral of \[tan(x)\;\] . If not, it's easy to find:
\[ \Rightarrow xtan(x) - \int {\dfrac{{\sin (x)}}{{\cos (x)}}} dx\]
Let \[t = cos(x)\] , implying that \[dt = - sin(x)dx\]
\[ \Rightarrow \smallint x{\sec ^2}(x)dx = x\tan (x) + \smallint \dfrac{{ - \sin (x)}}{{\cos (x)}}dx\]
Solving the equation, until it is simplified,
\[ \Rightarrow \smallint x{\sec ^2}(x)dx = x\tan (x) + \smallint \dfrac{1}{t}dt\]
This is a common integral:
\[ \Rightarrow \smallint x{\sec ^2}(x)dx = x\tan (x) + \ln \left| t \right| + C\]
Working back from \[t = cos(x)\]:
\[\smallint x{\sec ^2}(x)dx = x\tan (x) + \ln |\cos (x)| + C\;\]
Therefore, for an equation $\int {x\,{{\sec }^2}\,x} $ the integration by parts will make the solution come up to $x\tan (x) + \ln |\cos x| + C$
Note: Integration is a way of adding the slices to add the whole. It is the calculation of an integral. Integration by parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. Integration by parts is basically used for functions that can also be constructed as products of some other function and a third function’s derivative.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

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

When was the first election held in India a 194748 class 12 sst CBSE

