Answer
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Hint: Here, we will first use the Integration by Parts formula and rewrite the given integrand. We will then use the suitable integration formula to find the integral of the given function. Integration is defined as the summation of all the discrete data.
Formula used:
We will use the following formulas:
1. Integration by Parts: \[\int {uvdx = uv - \int {vdu} } \]
2. Derivative Formula: \[\dfrac{d}{{dx}}\left( C \right) = 1\]
3. Integral Formula: \[\int {{{\csc }^2}xdx = - \cot x} \], and\[\int {\cot xdx = \int {\dfrac{{\cos x}}{{\sin x}}dx} = \ln \left( {\sin x} \right) + C} \]
Complete Step by Step Solution:
We are given an Integral function \[\int {x{{\csc }^2}xdx} \].
Now, we will find the integral function by using Integration by Parts \[\int {uvdx = uv - \int {vdu} } \] for the given Integral function, we get \[u = x\] according to ILATE rule and \[v = {\csc ^2}x\].
Now, we will differentiate the variable\[u\], so we get
\[du = dx\]…………………………………………………….\[\left( 1 \right)\]
We will now integrate the variable \[v\] using the formula \[\int {{{\csc }^2}xdx = - \cot x} \], so we get
\[\int {{{\csc }^2}xdx = - \cot x} \] ……………………………………………………………………\[\left( 2 \right)\]
By substituting equation \[\left( 1 \right)\] and \[\left( 2 \right)\] in the integration by parts formula \[\int {uvdx = uv - \int {vdu} } \], we get
\[\int {x{{\csc }^2}xdx = - x\cot x - \int { - \cot xdx} } \] …………………………………………..\[\left( 3 \right)\]
Now, by rewriting the equation \[\left( 3 \right)\] in the integral function, we get
\[ \Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \int {\cot xdx} \]
Substituting \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] in the above equation, we get
\[ \Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \int {\dfrac{{\cos x}}{{\sin x}}dx} \]
Now, we will integrate the function by using the integral formula \[\int {\dfrac{{\cos x}}{{\sin x}}dx = \ln \left( {\sin x} \right) + C} \].
\[ \Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \left[ {\ln \left( {\sin x} \right)} \right] + C\]
Therefore, the value of \[\int {x{{\csc }^2}xdx} \] is \[ - x\cot x + \ln \left( {\sin x} \right) + C\].
Note:
We know that Integration is the process of adding small parts to find the whole parts. While performing the Integration by Parts, the first function is selected according to ILATE rule where Inverse Trigonometric function, followed by Logarithmic function, Arithmetic Function, Trigonometric Function and at last Exponential Function. Integration by Parts is applicable only when the integrand is a product of two Functions. Whenever the integration is done with no limits, then an Arbitrary constant should be added at the last step of the Integration.
Formula used:
We will use the following formulas:
1. Integration by Parts: \[\int {uvdx = uv - \int {vdu} } \]
2. Derivative Formula: \[\dfrac{d}{{dx}}\left( C \right) = 1\]
3. Integral Formula: \[\int {{{\csc }^2}xdx = - \cot x} \], and\[\int {\cot xdx = \int {\dfrac{{\cos x}}{{\sin x}}dx} = \ln \left( {\sin x} \right) + C} \]
Complete Step by Step Solution:
We are given an Integral function \[\int {x{{\csc }^2}xdx} \].
Now, we will find the integral function by using Integration by Parts \[\int {uvdx = uv - \int {vdu} } \] for the given Integral function, we get \[u = x\] according to ILATE rule and \[v = {\csc ^2}x\].
Now, we will differentiate the variable\[u\], so we get
\[du = dx\]…………………………………………………….\[\left( 1 \right)\]
We will now integrate the variable \[v\] using the formula \[\int {{{\csc }^2}xdx = - \cot x} \], so we get
\[\int {{{\csc }^2}xdx = - \cot x} \] ……………………………………………………………………\[\left( 2 \right)\]
By substituting equation \[\left( 1 \right)\] and \[\left( 2 \right)\] in the integration by parts formula \[\int {uvdx = uv - \int {vdu} } \], we get
\[\int {x{{\csc }^2}xdx = - x\cot x - \int { - \cot xdx} } \] …………………………………………..\[\left( 3 \right)\]
Now, by rewriting the equation \[\left( 3 \right)\] in the integral function, we get
\[ \Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \int {\cot xdx} \]
Substituting \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] in the above equation, we get
\[ \Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \int {\dfrac{{\cos x}}{{\sin x}}dx} \]
Now, we will integrate the function by using the integral formula \[\int {\dfrac{{\cos x}}{{\sin x}}dx = \ln \left( {\sin x} \right) + C} \].
\[ \Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \left[ {\ln \left( {\sin x} \right)} \right] + C\]
Therefore, the value of \[\int {x{{\csc }^2}xdx} \] is \[ - x\cot x + \ln \left( {\sin x} \right) + C\].
Note:
We know that Integration is the process of adding small parts to find the whole parts. While performing the Integration by Parts, the first function is selected according to ILATE rule where Inverse Trigonometric function, followed by Logarithmic function, Arithmetic Function, Trigonometric Function and at last Exponential Function. Integration by Parts is applicable only when the integrand is a product of two Functions. Whenever the integration is done with no limits, then an Arbitrary constant should be added at the last step of the Integration.
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