Algebra Questions

Question Number 145975 by mathdanisur last updated on 09/Jul/21

$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{6}} {\int}}\:\left[\:\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }−\left(\mathrm{6}−{x}\right)\right]{dx}=? \\$$

Answered by puissant last updated on 09/Jul/21

$$\mathrm{x}=\mathrm{6sin}\left(\mathrm{t}\right)\Rightarrow\mathrm{dx}=\mathrm{6cos}\left(\mathrm{t}\right)\mathrm{dt} \\$$$$\mathrm{I}=\mathrm{6}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{t}\right)}−\left(\mathrm{1}−\mathrm{sin}\left(\mathrm{t}\right)\mathrm{dt}\right. \\$$$$=\mathrm{6}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\mathrm{cos}\left(\mathrm{t}\right)−\mathrm{sin}\left(\mathrm{t}\right)−\mathrm{1}\right]\mathrm{dt} \\$$$$=\mathrm{6}\left[\mathrm{sin}\left(\mathrm{t}\right)+\mathrm{cos}\left(\mathrm{t}\right)−\mathrm{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\:\mathrm{6}\left[\left(\mathrm{1}−\frac{\pi}{\mathrm{2}}\right)−\mathrm{1}\right] \\$$$$\mathrm{I}=−\mathrm{3}\pi... \\$$

Commented by mathdanisur last updated on 10/Jul/21

$${thankyou}\:{Ser} \\$$