How To Integrate sqrt[-x^2+2x+5] by Trigonometric Substitution

Published On :2021-01-28 21:02:00

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begin{array}{l} smallint sqrt { - left( {{x^{bf{2}}} - {bf{2}}x - {bf{5}}} right)} {rm{d}}x,Notesqrt {a{x^2} + bx + c} Rightarrow byCompletingSquare\\ Rightarrow smallint sqrt { - left( {{x^2} - 2x + 1 - 1 - 5} right)} {rm{d}}x\ Rightarrow smallint sqrt {6 - {{left( {x - 1} right)}^2}} {rm{d}}x\ Let Rightarrow x - 1 = sqrt 6 sin theta \ {rm{d}}x = sqrt 6 cos theta dtheta \ Change{rm{ }}theexpression\ 6 - {left( {x - 1} right)^2} = 6left( {1 - {{sin }^2}theta } right)\ 6 - {left( {x - 1} right)^2} = 6{cos ^2}theta \ sqrt {6 - {{left( {x - 1} right)}^2}} = sqrt 6 cos theta \ smallint sqrt 6 cos theta sqrt 6 cos theta {rm{d}}theta = 6smallint co{s^2}theta dtheta \ 6smallint frac{{1 + cos 2theta }}{2}{rm{d}}theta = 3left( {theta + frac{{sin2theta }}{2}} right) + c\ Rightarrow 3theta + 3sin theta cos theta + c end{array}
begin{array}{l} Rightarrow 3{sin ^{ - 1}}(frac{{x - 1}}{{sqrt 6 }}) + 3 cdot left( {frac{{x - 1}}{{sqrt 6 }}} right) cdot left( {frac{{sqrt {6 - {{left( {x - 1} right)}^2}} }}{{sqrt 6 }}} right) + c end{array}