Question Number 145975 by mathdanisur last updated on 09/Jul/21
![∫_( 0) ^( 6) [ (√(36−x^2 ))−(6−x)]dx=?](Q145975.png)
$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{6}} {\int}}\:\left[\:\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }−\left(\mathrm{6}−{x}\right)\right]{dx}=? \\ $$
Answered by puissant last updated on 09/Jul/21
![x=6sin(t)⇒dx=6cos(t)dt I=6∫_0 ^(π/2) (√(1−sin^2 (t)))−(1−sin(t)dt =6∫_0 ^(π/2) [cos(t)−sin(t)−1]dt =6[sin(t)+cos(t)−t]_0 ^(π/2) = 6[(1−(π/2))−1] I=−3π...](Q145978.png)
$$\mathrm{x}=\mathrm{6sin}\left(\mathrm{t}\right)\Rightarrow\mathrm{dx}=\mathrm{6cos}\left(\mathrm{t}\right)\mathrm{dt} \\ $$ $$\mathrm{I}=\mathrm{6}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{t}\right)}−\left(\mathrm{1}−\mathrm{sin}\left(\mathrm{t}\right)\mathrm{dt}\right. \\ $$ $$=\mathrm{6}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\mathrm{cos}\left(\mathrm{t}\right)−\mathrm{sin}\left(\mathrm{t}\right)−\mathrm{1}\right]\mathrm{dt} \\ $$ $$=\mathrm{6}\left[\mathrm{sin}\left(\mathrm{t}\right)+\mathrm{cos}\left(\mathrm{t}\right)−\mathrm{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\:\mathrm{6}\left[\left(\mathrm{1}−\frac{\pi}{\mathrm{2}}\right)−\mathrm{1}\right] \\ $$ $$\mathrm{I}=−\mathrm{3}\pi... \\ $$
Commented bymathdanisur last updated on 10/Jul/21
$${thankyou}\:{Ser} \\ $$