Question

The value of $\int {{\text{x}}\left( {{\text{cosecx cotx}}} \right)} {\text{dx}}$= is
A. \[{\text{xcosecx - log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right|{\text{ + c}}\]
B. 2-\[{\text{xcosecx - log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right|{\text{ + c}}\]
C. \[{\text{xcosecx - 2log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right|{\text{ + c}}\]
D. \[{\text{xcotx - log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right|{\text{ + c}}\]

Answer 1

Hint: To solve this question, we need to apply the theory and various rules of integration. As we know, a given expression has two functions and we need to find the integration of multiplication of these two functions. So, we will apply the ‘By parts’ rule of integration. As we discussed below.

Complete step-by-step answer:

We know that if ‘u’ and ‘v’ are two functions then using ‘By parts’ rule, we can write:

where

‘u’ is taken as first function and

‘v’ is taken as a second function.

In this question, we have

$\int {{\text{x}}\left( {{\text{cosecx cotx}}} \right)} {\text{dx}}$

Here, first function is x and second function is $\left( {{\text{cosecx cotx}}} \right)$

 u= x and v= $\left( {{\text{cosecx cotx}}} \right)$

Now, apply ‘By parts’ rule,

$\int {{\text{x}}\left( {{\text{cosecx cotx}}} \right)} {\text{dx}}$= x $\int {{\text{cosecxcotxdx}}} - \int {\left( {\dfrac{{{\text{d(x)}}}}{{{\text{dx}}}} \times \int {{\text{cosecxcotxdx}}} } \right)} $dx

As we know,

 $\int {{\text{cosecxcotxdx}}} $= - cosecx +C

 $\int {{\text{ - cosecxdx}}} $= \[{\text{log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}}

\right|{\text{ + c}}\]

Now, using these integration values, we get

                                      = x (– cosec x) - $\int {{\text{ - cosecxdx}}} $

                                     = - xcosecx + \[{\text{log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}}

\right|{\text{ + c}}\]

As we know c is an arbitrary constant. So we take c as a C+2

So, = - xcosecx + \[{\text{log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right|{\text{ + C + 2}}\]

                                     = 2-xcosecx + \[{\text{log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}}

\right|{\text{ + C}}\]

Therefore, the value of $\int {{\text{x}}\left( {{\text{cosecx cotx}}} \right)} {\text{dx}}$ is 2-xcosec x + \[{\text{log}}\left| {{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right|{\text{ + C}}\].

So, option (B) is the correct answer.

Note: In this question, the important step is division of a given function into two parts. You should remember the method of solving integration using the ‘By part’ rule. Here the first and second function are generally chosen according to ILATE rule, where:

I- Inverse function
L- Logarithmic function
A- Algebraic function
T- Trigonometric function
E- Exponential function
Second function is chosen in decreasing order from top to bottom.

Subject : Mathematics