Question Number 13960 by tawa tawa last updated on 25/May/17
$$\int_{\:\:\:\mathrm{0}} ^{\:\frac{\mathrm{a}}{\mathrm{2}}} \:\:\:\mathrm{x}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} \:−\:\mathrm{x}^{\mathrm{2}} \right)^{−\mathrm{3}/\mathrm{2}} \:\:\mathrm{dx} \\ $$
Answered by ajfour last updated on 25/May/17
![I=∫_0 ^( a/2) (x^2 /((a^2 −x^2 )^(3/2) ))dx let x=asin θ ⇒ dx=acos θdθ when x=0, θ=0 when x=(a/2), θ=(π/6) I= ∫_0 ^( π/6) ((a^2 sin^2 θ)/(a^3 cos^3 θ))(acos θdθ) = ∫_0 ^( π/6) tan^2 θdθ = ∫_0 ^( π/6) (sec^2 θ−1)dθ = [tan θ−θ]∣_0 ^(π/6) I= (1/( (√3)))−(π/6) =(((2(√3)−π))/6) .](https://www.tinkutara.com/question/Q13963.png)
$${I}=\int_{\mathrm{0}} ^{\:{a}/\mathrm{2}} \frac{{x}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }{dx} \\ $$$${let}\:{x}={a}\mathrm{sin}\:\theta\:\:\:\Rightarrow\:\:{dx}={a}\mathrm{cos}\:\theta{d}\theta \\ $$$${when}\:{x}=\mathrm{0},\:\theta=\mathrm{0} \\ $$$${when}\:{x}=\frac{{a}}{\mathrm{2}},\:\theta=\frac{\pi}{\mathrm{6}} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\:\pi/\mathrm{6}} \:\frac{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{a}^{\mathrm{3}} \mathrm{cos}\:^{\mathrm{3}} \theta}\left({a}\mathrm{cos}\:\theta{d}\theta\right) \\ $$$$\:\:=\:\int_{\mathrm{0}} ^{\:\pi/\mathrm{6}} \mathrm{tan}\:^{\mathrm{2}} \theta{d}\theta \\ $$$$\:\:=\:\int_{\mathrm{0}} ^{\:\pi/\mathrm{6}} \left(\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{1}\right){d}\theta \\ $$$$\:\:=\:\left[\mathrm{tan}\:\theta−\theta\right]\mid_{\mathrm{0}} ^{\pi/\mathrm{6}} \\ $$$${I}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\frac{\pi}{\mathrm{6}}\:\:=\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}−\pi\right)}{\mathrm{6}}\:. \\ $$
Commented by tawa tawa last updated on 25/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{understand}\:\mathrm{now}. \\ $$