Let #160;I= #8747;x1+x x2dx #38; #160;let #160;x= #160;Addx1+x x2+B #8658;x=A #160; 2x+1+BBy #160;equating #160;the #160;coefficients #1 ...

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Integration of Trigonometric Functions

∫ x 1+x-x 2 d...

Question

$\int x \sqrt{1 + x - x^{2}} d x$

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Solution

$Let I = \int x \sqrt{1 + x - x^{2}} d x & let x = A \frac{d}{d x} (1 + x - x^{2}) + B \Rightarrow x = A (- 2 x + 1) + B By equating the coefficients of like terms we get, x = (- 2 A) x \Rightarrow A = - \frac{1}{2} & A + B = 0 \Rightarrow B = \frac{1}{2} By substituting the values of A and B in eq (1) we get, I = \int [- \frac{1}{2} (- 2 x + 1) + \frac{1}{2}] \sqrt{1 + x - x^{2}} d x = - \frac{1}{2} \int (- 2 x + 1) \sqrt{1 + x - x^{2}} d x + \frac{1}{2} \sqrt{1 + x - x^{2}} d x Putting 1 + x - x^{2} = t \Rightarrow (- 2 x + 1) d x = d t ∴ I = - \frac{1}{2} \int \sqrt{t} \cdot d t + \frac{1}{2} \int \sqrt{1 + x - x^{2}} d x = - \frac{1}{2} \int \sqrt{t} d t + \frac{1}{2} \int \sqrt{1 - (x^{2} - x)} d x = - \frac{1}{2} \int t^{\frac{1}{2}} \cdot d t + \frac{1}{2} \int \sqrt{1 - (x^{2} - x + \frac{1}{4} - \frac{1}{4})} d x = - \frac{1}{2} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + \frac{1}{2} \int \sqrt{1 - {(x - \frac{1}{2})}^{2} + \frac{1}{4}} d x = - \frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}} + \frac{1}{2} \int \sqrt{{(\frac{\sqrt{5}}{2})}^{2} - {(x - \frac{1}{2})}^{2}} d x = - \frac{1}{3} t^{\frac{3}{2}} + \frac{1}{2} [(\frac{x - \frac{1}{2}}{2}) \sqrt{{(\frac{\sqrt{5}}{2})}^{2} - {(x - \frac{1}{2})}^{2}} + \frac{{(\frac{\sqrt{5}}{2})}^{2}}{2} \sin^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{5}}{2}})] + C [∵ \int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{1}{2} a^{2} \sin^{- 1} \frac{x}{a} + C] = - \frac{1}{3} {(1 + x - x^{2})}^{\frac{3}{2}} + \frac{1}{2} [(\frac{2 x - 1}{4}) \sqrt{1 + x - x^{2}} + \frac{5}{8} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}})] + C = \frac{- (1 + x - x^{2}) \sqrt{1 + x - x^{2}}}{3} + \frac{(2 x - 1)}{8} \sqrt{1 + x - x^{2}} + \frac{5}{16} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}}) + C = \sqrt{1 + x - x^{2}} [\frac{- (1 + x - x^{2})}{3} + \frac{2 x - 1}{8}] + \frac{5}{16} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}}) + C = \sqrt{1 + x - x^{2}} [\frac{- 8 - 8 x + 8 x^{2} + 6 x - 3}{24}] + \frac{5}{16} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}}) + C = \sqrt{1 + x - x^{2}} [\frac{8 x^{2} - 2 x - 11}{24}] + \frac{5}{16} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}}) + C$

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