Evaluating the Integral of x * √(b2 - (a - x)2) Using Trigonometric Substitutions

In this article, we will explore the evaluation of a specific integral using trigonometric substitution. The integral in question is:

[ mathcal{I}(x) int x sqrt{b^2 - (a - x)^2} , dx ]

Let's break down the solution step by step, using a range of substitutions to simplify and evaluate the integral.

Substitution 1: Let ( x a - u )

We start by letting ( x a - u ). This substitution helps to simplify the expression inside the square root:

Substituting ( x a - u ) into the integral, we get: [ mathcal{I}(x) -int (a - u) sqrt{b^2 - u^2} , du ] Next, we further simplify the integral by letting ( u b sin(s) ): [ mathcal{I}(x) -b^2 int (a - b sin(s)) sqrt{1 - sin^2(s)} , cos(s) , ds ] Using ( sqrt{1 - sin^2(s)} cos(s) ): [ mathcal{I}(x) -b^2 int (a - b sin(s)) cos^2(s) , ds ]

Separating the Integral

We can split the integral into two parts:

[ mathcal{I}(x) -b^2 left[ a int cos^2(s) , ds - b int sin(s) cos^2(s) , ds right] ]

Evaluating ( mathcal{P}(s) int cos^2(s) , ds )

The first part of the integral involves ( int cos^2(s) , ds ). We use integration by parts to solve this:

[ int cos^2(s) , ds cos(s) sin(s) - int sin(s) cos(s) , ds ] Note that ( int sin(s) cos(s) , ds frac{1}{2} sin^2(s) ): Thus, ( int cos^2(s) , ds frac{1}{2} cos(s) sin(s) frac{1}{2} int cos(s) sin(s) , ds ) Let ( s arcsin(t) ) and ( t 1 - w ): [ int cos^2(s) , ds frac{1}{2} s sqrt{1 - sin^2(s)} frac{1}{2} arcsin(s) ] Combining these results, we get: [ mathcal{P}(s) frac{1}{2} s sqrt{1 - s^2} frac{1}{2} arcsin(s) C_{mathcal{P}} ]

Evaluating ( mathcal{Q}(s) int 2s cos^2(s) , ds )

The second part of the integral involves ( int 2s cos^2(s) , ds ). Using the same substitution ( s arcsin(t) ) and ( t 1 - w ), we get:

[ int 2s cos^2(s) , ds - int sqrt{w} , dw ] Integrating ( int sqrt{w} , dw frac{2}{3} w^{3/2} ): Thus, ( int 2s cos^2(s) , ds - frac{2}{3} (1 - s^2)^{3/2} C_{mathcal{Q}} )

Combining Results

Substituting the results of ( mathcal{P}(s) ) and ( mathcal{Q}(s) ) back into the original integral:

[ mathcal{I}(x) -left[ a b^2 left( frac{1}{2} s sqrt{1 - s^2} frac{1}{2} arcsin(s) right) - frac{2}{3} b^3 (1 - s^2)^{3/2} right] ] Undoing the substitutions ( s sin(u/b) ) and ( u b(x - a) ) gives: [ mathcal{I}(x) -left[ a b^2 left( frac{1}{2} frac{x - a}{b} sqrt{1 - left( frac{x - a}{b} right)^2} frac{1}{2} arcsin left( frac{x - a}{b} right) right) - frac{2}{3} b^3 left( 1 - left( frac{x - a}{b} right)^2 right)^{3/2} right] ] Simplifying the expression with ( a -1 ) and ( b 2 ): [ mathcal{I}(x) -left[ frac{1}{3} left( 4 (x 1) sqrt{4 - (x 1)^2} - (x 1)^2 sqrt{4 - (x 1)^2} right) - frac{1}{2} (-1)(4)(arcsin frac{x 1}{2}) right] ] Therefore, the final result is: [ mathcal{I}(x) -frac{1}{3} (4(x 1) - (x 1)^2) sqrt{4 - (x 1)^2} frac{1}{2} (x 1) arcsin frac{x 1}{2} ]