
Find the value of the integral \[\int {{e^x}\left( {{\text{cosecx}}} \right)} \left( {1 - \cot x} \right)dx\].
$
{\text{A}}{\text{. }}{{\text{e}}^x}\left( {\cot x} \right) + c \\
{\text{B}}{\text{. }}{{\text{e}}^x}\left( {{\text{cosecx}}} \right)\left( {\cot x} \right) + c \\
{\text{C}}{\text{. }}{{\text{e}}^x}\left( {{\text{cosecx}}} \right) + c \\
{\text{D}}{\text{. }} - {{\text{e}}^x}\left( {\cot x} \right) + c \\
$
Answer
216k+ views
Hint- Here, we will proceed by using the concept of integration by parts to solve the integral \[\int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx\] separately and then using the formula \[\int {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} dx = - \cos {\text{ec}}x\] to simplify the given integral whose value is required.
Complete step by step answer:
Let us then suppose the given integral be
\[
{\text{I}} = \int {{e^x}\left( {{\text{cosecx}}} \right)} \left( {1 - \cot x} \right)dx \\
\Rightarrow {\text{I}} = \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right) - {e^x}\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx \\
\Rightarrow {\text{I}} = \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)} \right]} dx - \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx{\text{ }} \to {\text{(1)}} \\
\]
According to the formula of integration by parts (ILATE), the integral consisting of two different functions in variable x can be evaluated as under
\[\int {{\text{f}}\left( x \right)g\left( x \right)} dx = {\text{f}}\left( x \right)\left[ {\int {g\left( x \right)} dx} \right] - \int {\left\{ {\left[ {\dfrac{d}{{dx}}\left[ {{\text{f}}\left( x \right)} \right]} \right]\int {g\left( x \right)dx} } \right\}} dx + c{\text{ }} \to {\text{(2)}}\] where c is any constant of integration
where the functions \[{\text{f}}\left( x \right)\] and \[g\left( x \right)\] are arranged based on the priority basis according to ILATE
Considering the integral \[\int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx\] in equation (1) where the first function is \[f\left( x \right) = {e^x}\] and the second function is \[g\left( x \right) = \left( {{\text{cosecx}}} \right)\left( {\cot x} \right)\] and then solving it using the formula given by equation (2), we get
\[ \Rightarrow \int {{e^x}\left[ {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx = {e^x}\left[ {\int {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} dx} \right] - \int {\left\{ {\left[ {\dfrac{d}{{dx}}\left( {{e^x}} \right)} \right]\left[ {\int {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} dx} \right]} \right\}} dx + c\]
Using the formula \[\int {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} dx = - \cos {\text{ec}}x\], the above integral becomes
\[
\Rightarrow \int {{e^x}\left[ {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx = {e^x}\left[ { - \cos {\text{ec}}x} \right] - \int {\left\{ {\left[ {{e^x}} \right]\left[ { - \cos {\text{ec}}x} \right]} \right\}} dx + c \\
\Rightarrow \int {{e^x}\left[ {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx = - {e^x}\left( {\cos {\text{ec}}x} \right) + \int {\left[ {{e^x}\left( {\cos {\text{ec}}x} \right)} \right]} dx + c{\text{ }} \to {\text{(3)}} \\
\]
By substituting equation (3) in equation (1), we get
\[
\Rightarrow {\text{I}} = \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)} \right]} dx - \left[ { - {e^x}\left( {\cos {\text{ec}}x} \right) + \int {\left[ {{e^x}\left( {\cos {\text{ec}}x} \right)} \right]} dx} \right] + c \\
\Rightarrow {\text{I}} = \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)} \right]} dx + {e^x}\left( {\cos {\text{ec}}x} \right) - \int {\left[ {{e^x}\left( {\cos {\text{ec}}x} \right)} \right]} dx + c \\
\Rightarrow {\text{I}} = {e^x}\left( {\cos {\text{ec}}x} \right) + c \\
\]
Therefore, the integral \[\int {{e^x}\left( {{\text{cosecx}}} \right)} \left( {1 - \cot x} \right)dx = {e^x}\left( {\cos {\text{ec}}x} \right) + c\]
Hence, option C is correct.
Note- In the method of integration by parts (ILATE), I refers to inverse trigonometric function, L refers to logarithmic function, A refers to algebraic function, T refers to trigonometric function and E refers to exponential function. In this problem, the first function is taken as ${e^x}$ and the second function as \[\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)\] so that the required integral can be solved conveniently.
Complete step by step answer:
Let us then suppose the given integral be
\[
{\text{I}} = \int {{e^x}\left( {{\text{cosecx}}} \right)} \left( {1 - \cot x} \right)dx \\
\Rightarrow {\text{I}} = \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right) - {e^x}\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx \\
\Rightarrow {\text{I}} = \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)} \right]} dx - \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx{\text{ }} \to {\text{(1)}} \\
\]
According to the formula of integration by parts (ILATE), the integral consisting of two different functions in variable x can be evaluated as under
\[\int {{\text{f}}\left( x \right)g\left( x \right)} dx = {\text{f}}\left( x \right)\left[ {\int {g\left( x \right)} dx} \right] - \int {\left\{ {\left[ {\dfrac{d}{{dx}}\left[ {{\text{f}}\left( x \right)} \right]} \right]\int {g\left( x \right)dx} } \right\}} dx + c{\text{ }} \to {\text{(2)}}\] where c is any constant of integration
where the functions \[{\text{f}}\left( x \right)\] and \[g\left( x \right)\] are arranged based on the priority basis according to ILATE
Considering the integral \[\int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx\] in equation (1) where the first function is \[f\left( x \right) = {e^x}\] and the second function is \[g\left( x \right) = \left( {{\text{cosecx}}} \right)\left( {\cot x} \right)\] and then solving it using the formula given by equation (2), we get
\[ \Rightarrow \int {{e^x}\left[ {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx = {e^x}\left[ {\int {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} dx} \right] - \int {\left\{ {\left[ {\dfrac{d}{{dx}}\left( {{e^x}} \right)} \right]\left[ {\int {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} dx} \right]} \right\}} dx + c\]
Using the formula \[\int {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} dx = - \cos {\text{ec}}x\], the above integral becomes
\[
\Rightarrow \int {{e^x}\left[ {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx = {e^x}\left[ { - \cos {\text{ec}}x} \right] - \int {\left\{ {\left[ {{e^x}} \right]\left[ { - \cos {\text{ec}}x} \right]} \right\}} dx + c \\
\Rightarrow \int {{e^x}\left[ {\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)} \right]} dx = - {e^x}\left( {\cos {\text{ec}}x} \right) + \int {\left[ {{e^x}\left( {\cos {\text{ec}}x} \right)} \right]} dx + c{\text{ }} \to {\text{(3)}} \\
\]
By substituting equation (3) in equation (1), we get
\[
\Rightarrow {\text{I}} = \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)} \right]} dx - \left[ { - {e^x}\left( {\cos {\text{ec}}x} \right) + \int {\left[ {{e^x}\left( {\cos {\text{ec}}x} \right)} \right]} dx} \right] + c \\
\Rightarrow {\text{I}} = \int {\left[ {{e^x}\left( {{\text{cosecx}}} \right)} \right]} dx + {e^x}\left( {\cos {\text{ec}}x} \right) - \int {\left[ {{e^x}\left( {\cos {\text{ec}}x} \right)} \right]} dx + c \\
\Rightarrow {\text{I}} = {e^x}\left( {\cos {\text{ec}}x} \right) + c \\
\]
Therefore, the integral \[\int {{e^x}\left( {{\text{cosecx}}} \right)} \left( {1 - \cot x} \right)dx = {e^x}\left( {\cos {\text{ec}}x} \right) + c\]
Hence, option C is correct.
Note- In the method of integration by parts (ILATE), I refers to inverse trigonometric function, L refers to logarithmic function, A refers to algebraic function, T refers to trigonometric function and E refers to exponential function. In this problem, the first function is taken as ${e^x}$ and the second function as \[\left( {{\text{cosecx}}} \right)\left( {\cot x} \right)\] so that the required integral can be solved conveniently.
Recently Updated Pages
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Equation of Trajectory in Projectile Motion: Derivation & Proof

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Collision: Meaning, Types & Examples in Physics

How to Convert a Galvanometer into an Ammeter or Voltmeter

Atomic Structure: Definition, Models, and Examples

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Average and RMS Value in Physics: Formula, Comparison & Application

