Question Number 20386 by tammi last updated on 26/Aug/17
$$\int\sqrt{\mathrm{1}−{a}^{\mathrm{2}} {x}^{\mathrm{2}} {dx}} \\ $$
Answered by $@ty@m last updated on 26/Aug/17
![Let x=((sinθ)/a) ⇒dx=(1/a)cosθdθ ⇒I=(1/a)∫(√(1−sin^2 θ))cosθdθ =(1/a)∫cos^2 θdθ =(1/(2a))∫(1+cos2θ)dθ =(1/(2a))[θ+((sin2θ)/2)]+C =(1/(2a))[sin^(−1) ax+ax(√(1−a^2 x^2 ))]+C](https://www.tinkutara.com/question/Q20406.png)
$${Let}\:{x}=\frac{{sin}\theta}{{a}} \\ $$$$\Rightarrow{dx}=\frac{\mathrm{1}}{{a}}{cos}\theta{d}\theta \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{{a}}\int\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta}{cos}\theta{d}\theta \\ $$$$=\frac{\mathrm{1}}{{a}}\int{cos}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}}\int\left(\mathrm{1}+{cos}\mathrm{2}\theta\right){d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}}\left[\theta+\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right]+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}}\left[{sin}^{−\mathrm{1}} {ax}+{ax}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} {x}^{\mathrm{2}} }\right]+{C} \\ $$