Question Number 116016 by bemath last updated on 30/Sep/20
$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\:\sqrt{\mathrm{6}+{x}−{x}^{\mathrm{2}} }}\:? \\ $$
Answered by bobhans last updated on 30/Sep/20
![I=∫_(−1) ^1 (dx/( (√(6−(x^2 −x))))) = ∫_(−1) ^1 (dx/( (√(((25)/4)−(x−(1/2))^2 )))) I= [ sin^(−1) (((x−(1/2))/(5/2))) ]_(−1) ^1 =[ sin^(−1) (((2x−1)/5)) ]_(−1) ^1 = sin^(−1) (0.2)+sin^(−1) (0.6)](https://www.tinkutara.com/question/Q116017.png)
$${I}=\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\:\sqrt{\mathrm{6}−\left({x}^{\mathrm{2}} −{x}\right)}}\:=\:\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\:\sqrt{\frac{\mathrm{25}}{\mathrm{4}}−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$${I}=\:\left[\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{5}}{\mathrm{2}}}\right)\:\right]_{−\mathrm{1}} ^{\mathrm{1}} =\left[\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{5}}\right)\:\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{2}\right)+\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{6}\right) \\ $$