
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
244.5k+ 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.
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