
What is the formula for integration of \[uv\]?
Answer
161.1k+ views
Hint: We use the concept of integration to get the formula for integration. First of all use the product rule of derivatives and then rearrange the equation and integrate the equation to get the formula for integration of \[uv\]. The integration indicates the summation of discrete data.
Complete step-by-step solution:
Integrating the \[uv\] formula is a convenient way to find the integral of the product of two functions. Also, the two functions used in this integration of \[uv\] expression are algebraic expressions, trigonometric functions, and logarithmic functions. Extends the derivative of the product of functions to represent the integral given by a known integral. Therefore, the integral of the \[uv\]expression is also called the integration by parts or the product rule of integrals. Integration by parts is not applicable to functions like \[\int \sqrt{ x }\sin x dx\]
If \[u\]and \[v\] are two functions then the integral of the \[uv\] expression is given as follows:
\[\int {uv\,dx = u\int {v\,dx - \int {\left( {\dfrac{{du}}{{dx}}\int {v\,dx} } \right)} } } \,dx\]
Where u is the first function and v is the second function. To choose the first function we take the help of the ILATE rule.
Proof:
Use the product rule to derive the integral of the \[uv\]equation. Let two functions v and u such that
\[y = uv\].Apply the following product rule.
\[\dfrac{d}{{dx}}(uv) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}\]
Now rearrange the following equation
\[u\dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}(uv) - v\dfrac{{du}}{{dx}}\]
Now on integrating both sides with respect x we get,
\[\int {\left( {u\dfrac{{dv}}{{dx}}} \right)} dx = \int {\left( {\dfrac{d}{{dx}}(uv) - v\dfrac{{du}}{{dx}}} \right)} dx\]
\[\int {\left( {u\dfrac{{dv}}{{dx}}} \right)} dx = \int {\dfrac{d}{{dx}}(uv)dx - v\dfrac{{du}}{{dx}}} dx\]
\[\]\[ \Rightarrow \int u\,dv = {\rm{ uv }} - {\rm{ }}\int v{\rm{ du}}\]
To find the value of \[\int vu'dx\] we need to find the indefinite integral of \[v'\]that exists in the original integral \[\int uv'dx.\]
Note: The integration by part follows a rule to choose the first function and second function. To choose the first term, we used the formula ILATE formula.
I \[\rightarrow\] Inverse trigonometry function
L \[\rightarrow\] Logarithm function
A \[\rightarrow\] Algeberical expression
T \[\rightarrow\] Trigonometry ratios
E \[\rightarrow\] Exponential function
Complete step-by-step solution:
Integrating the \[uv\] formula is a convenient way to find the integral of the product of two functions. Also, the two functions used in this integration of \[uv\] expression are algebraic expressions, trigonometric functions, and logarithmic functions. Extends the derivative of the product of functions to represent the integral given by a known integral. Therefore, the integral of the \[uv\]expression is also called the integration by parts or the product rule of integrals. Integration by parts is not applicable to functions like \[\int \sqrt{ x }\sin x dx\]
If \[u\]and \[v\] are two functions then the integral of the \[uv\] expression is given as follows:
\[\int {uv\,dx = u\int {v\,dx - \int {\left( {\dfrac{{du}}{{dx}}\int {v\,dx} } \right)} } } \,dx\]
Where u is the first function and v is the second function. To choose the first function we take the help of the ILATE rule.
Proof:
Use the product rule to derive the integral of the \[uv\]equation. Let two functions v and u such that
\[y = uv\].Apply the following product rule.
\[\dfrac{d}{{dx}}(uv) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}\]
Now rearrange the following equation
\[u\dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}(uv) - v\dfrac{{du}}{{dx}}\]
Now on integrating both sides with respect x we get,
\[\int {\left( {u\dfrac{{dv}}{{dx}}} \right)} dx = \int {\left( {\dfrac{d}{{dx}}(uv) - v\dfrac{{du}}{{dx}}} \right)} dx\]
\[\int {\left( {u\dfrac{{dv}}{{dx}}} \right)} dx = \int {\dfrac{d}{{dx}}(uv)dx - v\dfrac{{du}}{{dx}}} dx\]
\[\]\[ \Rightarrow \int u\,dv = {\rm{ uv }} - {\rm{ }}\int v{\rm{ du}}\]
To find the value of \[\int vu'dx\] we need to find the indefinite integral of \[v'\]that exists in the original integral \[\int uv'dx.\]
Note: The integration by part follows a rule to choose the first function and second function. To choose the first term, we used the formula ILATE formula.
I \[\rightarrow\] Inverse trigonometry function
L \[\rightarrow\] Logarithm function
A \[\rightarrow\] Algeberical expression
T \[\rightarrow\] Trigonometry ratios
E \[\rightarrow\] Exponential function
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