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

$$\underset{0}{\stackrel{6}{\int }}[\sqrt{36-x^2}-(6-x)]dx=?$$

Answered by puissant last updated on 09/Jul/21

$$\mathrm{x}=6\sin \left(\mathrm{t}\right)\Rightarrow \mathrm{dx}=6\cos \left(\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{I}=6{\int }_0^{\frac{\pi }{2}}\sqrt{1-{\sin }^2\left(\mathrm{t}\right)}-(1-\sin \left(\mathrm{t}\right)\mathrm{dt}$$$$=6{\int }_0^{\frac{\pi }{2}}[\cos \left(\mathrm{t}\right)-\sin \left(\mathrm{t}\right)-1]\mathrm{dt}$$$$=6{[\sin \left(\mathrm{t}\right)+\cos \left(\mathrm{t}\right)-\mathrm{t}]}_0^{\frac{\pi }{2}}=6[(1-\frac{\pi }{2})-1]$$$$\mathrm{I}=-3\pi ...$$

Commented by mathdanisur last updated on 10/Jul/21

$$thankyouSer$$

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