Question Number 211184 by nothing48 last updated on 30/Aug/24
Answered by mm1342 last updated on 30/Aug/24
$${x}={sinu} \\ $$$$\Rightarrow\int{sin}^{\mathrm{3}} {udu}=\int\left(\mathrm{1}−{cos}^{\mathrm{2}} {u}\right){sinudu} \\ $$$$=−{cosu}+\frac{\mathrm{1}}{\mathrm{3}}{cos}^{\mathrm{3}} {u}+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}−{cos}^{\mathrm{2}} {u}\right){cosu}+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{c}\:\:\checkmark \\ $$$$ \\ $$
Answered by Frix last updated on 30/Aug/24
$$\int\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:\overset{{t}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {=}\:\int\left({t}^{\mathrm{2}} −\mathrm{1}\right){dt}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}= \\ $$$$=\left({t}^{\mathrm{2}} −\mathrm{3}\right)\frac{{t}}{\mathrm{3}}=−\left({x}^{\mathrm{2}} +\mathrm{2}\right)\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}+{C} \\ $$
Commented by mm1342 last updated on 30/Aug/24
$$\:\underbrace{\frown} \\ $$
Commented by Frix last updated on 30/Aug/24
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