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pradyot mayank Grade: 12
        

what is newton -leibnitz's rule how to apply it in a question ?explain with 3-4 example

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear pradyot


Leibniz Integral Rule


The Leibniz integral rule gives a formula for differentiation of a definite integral whose limits are functions of the differential variable,




5849-972_7662_908735170818779a3a888ae684ae94d3.png


 


where the partial derivative of f indicates that inside the integral only the variation of ƒ ( x, α ) with α is considered in taking the derivative.


Example


Here, we consider the integration of


\textbf I\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;dx,\,

where both a,\,b\,>\,0, by differentiating under the integral sign.


Let us first find \textbf J\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{a\,\cos^2\,x+b\,\sin^2\,x}\;dx.\,


Dividing both the numerator and the denominator by \cos^2\,x yields


  \begin{align}     \textbf J\;     &=\;\int_0^{\frac{\pi}{2}}\,\frac{\sec^2\,x}{a\,+b\,\tan^2\,x}\;dx     \\     &=\,\frac{1}{b}\,\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(\sqrt{\,\frac{a}{b}\,}\right)^2+\tan^2\,x}\;d(\tan\,x)\,     \\     &=\,\frac{1}{\sqrt{\,a\,b\,}}\,\left(\tan^{-1}\left(\sqrt{\,\frac{b}{a}\,}\,\tan\,x\right)\right)\,\bigg|_0^{\frac{\pi}{2}}\;=\;\frac{\pi}{2\,\sqrt{\,a\,b\,}}.   \end{align}

The limits of integration being independent of a,\, \textbf J\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{a\,\cos^2\,x+b\,\sin^2\,x}\;dx\, gives us


\frac{\partial\,\textbf J}{\partial\,a}\;=\;-\,\int_0^{\frac{\pi}{2}}\,\frac{\cos^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\,

whereas \textbf J\;=\;\frac{\pi}{2\,\sqrt{\,a\,b\,}} gives us


\frac{\partial\,\textbf J}{\partial\,a}\;=\;-\frac{\pi}{4\,\sqrt{\,a^3\,b\,}}.\,

Equating these two relations then yields


\,\int_0^{\frac{\pi}{2}}\,\frac{\cos^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;=\;\frac{\pi}{4\,\sqrt{\,a^3\,b\,}}.\,

In a similar fashion, pursuing \frac{\partial\,\textbf J}{\partial\,b}\, yields


\,\int_0^{\frac{\pi}{2}}\,\frac{\sin^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;=\;\frac{\pi}{4\,\sqrt{\,a\,b^3\,}}.\,

Adding the two results then produces


\textbf I\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;dx\;=\;\frac{\pi}{4\,\sqrt{\,a\,b\,}}\left(\frac{1}{a}+\frac{1}{b}\right),\,

which is the value of the integral \textbf I.\,


 





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Badiuddin

7 years ago
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