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Given ∫ X 0 1 X 2 + a 2 D X = 1 a Tan − 1 ( X a ) Using Duis Find the Value of ∫ X 0 1 X 2 + a 2 - Applied Mathematics 2

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प्रश्न

Given `int_0^x 1/(x^2+a^2) dx=1/atan^(-1)(x/a)`using DUIS find the value of `int_0^x 1/(x^2+a^2) `

योग

उत्तर

`int_0^x 1/(x^2+a^2) dx=1/atan^(-1)(x/a)`

Differentiate w.r.t a , taking ‘a’ as parameter

`d/(da)int_0^x 1/(x^2+a^2) dx=d/(da)[1/atan^(-1)(x/a)]`

Applying D.U.I.S rule,

D.U.I.S rule says that if function and its partial derivative is continuous then we can apply differential operator in the integral operator by converting it into partial derivative taking one parameter fro function.

`int_0^xdel/(dela) 1/(x^2+a^2) dx=-1/atan^(-1)(x/a)xx1/a+(-x)/(a(x^2+a^2)`

`int_0^x(2a^2)/(x^2+a^2) dx=-1/atan^(-1)(x/a)xx1/a+(-x)/(a(x^2+a^2)`

`int_0^x(dx)/(x^2+a^2)^2 dx=-1/(2a^3)tan^(-1)  x/a+x/(2a^2(x^2+a^2)`

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Method of Variation of Parameters
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