Let #160;I= #8747; #160;2x 5 #160;2+3x x2 #160;dxAlso, #160;2x 5= #955;ddx2+3x x2+ #956; #8658;2x 5= #955; 2x+3+ #956; #8658;2x 5= 2 # ...

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Integration of Trigonometric Functions

∫ 2 x-52+3 x-...

Question

$\int (2 x - 5) \sqrt{2 + 3 x - x^{2}} d x$

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Solution

$Let I = \int (2 x - 5) \sqrt{2 + 3 x - x^{2}} d x Also, 2 x - 5 = λ \frac{d}{d x} (2 + 3 x - x^{2}) + μ \Rightarrow 2 x - 5 = λ (- 2 x + 3) + μ \Rightarrow 2 x - 5 = (- 2 λ) x + 3 λ + μ Equating coeffieicents of like terms - 2 λ = 2 \Rightarrow λ = - 1 And 3 λ + μ = - 5 \Rightarrow 3 (- 1) + μ = - 5 \Rightarrow μ = - 5 + 3 \Rightarrow μ = - 2 ∴ 2 x - 5 = - 1 (- 2 x + 3) - 2 Hence, I = \int [- (- 2 x + 3) - 2] \sqrt{2 + 3 x - x^{2}} d x = - \int (- 2 x + 3) \sqrt{2 + 3 x - x^{2}} d x - 2 \int \sqrt{2 + 3 x - x^{2}} d x = - I_{1} - 2 I_{2} . . . . . (1) I_{1} = \int (- 2 x + 3) \sqrt{2 + 3 x - x^{2}} d x Let 2 + 3 x - x^{2} = t \Rightarrow (- 2 x + 3) d x = d t ∴ I_{1} = \int t^{\frac{1}{2}} d t = \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} = \frac{2}{3} t^{\frac{3}{2}} = \frac{2}{3} {(2 + 3 x - x^{2})}^{\frac{3}{2}} . . . . . (2) And I_{2} = \int \sqrt{2 + 3 x - x^{2}} d x I_{2} = \int \sqrt{2 - (x^{2} - 3 x)} d x = \int \sqrt{2 - [x^{2} - 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}]} d x = \int \sqrt{2 + \frac{9}{4} - {(x - \frac{3}{2})}^{2}} d x = \int \sqrt{{(\frac{\sqrt{17}}{2})}^{2} - {(x - \frac{3}{2})}^{2}} d x = \frac{x - \frac{3}{2}}{2} \sqrt{{(\frac{\sqrt{17}}{2})}^{2} - {(x - \frac{3}{2})}^{2}} + \frac{{(\frac{\sqrt{17}}{2})}^{2}}{2} \sin^{- 1} (\frac{x - \frac{3}{2}}{\frac{\sqrt{17}}{2}}) = \frac{2 x - 3}{4} \sqrt{2 + 3 x - x^{2}} + \frac{17}{8} \sin^{- 1} (\frac{2 x - 3}{\sqrt{17}}) . . . . . (3) From eq (1), (2) and (3) we have I = - \frac{2}{3} {(2 + 3 x - x^{2})}^{\frac{3}{2}} - \frac{(2 x - 3)}{2} \sqrt{2 + 3 x - x^{2}} - \frac{17}{4} \sin^{- 1} (\frac{2 x - 3}{\sqrt{17}}) + C$

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