Question Number 40147 by maxmathsup by imad last updated on 16/Jul/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\sqrt{{x}^{\mathrm{3}} \left(\mathrm{2}−{x}\right)}{dx} \\ $$
Commented by math khazana by abdo last updated on 17/Jul/18
![let I = ∫_0 ^2 (√(x^3 (2−x)))dx I =∫_0 ^2 x(√(x(2−x))) changement x=2sin^2 θ give I = ∫_0 ^(π/2) 2 sin^2 θ(√(2sin^2 θ(2−2sin^2 θ))) 4 sinθ cosθdθ =16∫_0 ^(π/2) sin^2 θ sinθ cosθ sinθ cosθ dθ =16 ∫_0 ^(π/2) sin^4 θ cos^2 θdθdθ by parts ∫_0 ^(π/2) cosθ( cosθ sin^4 θ)dθ =[(1/5) sin^5 θ cosθ]_0 ^(π/2) +∫_0 ^(π/2) (1/5) sin^5 θ sinθ dθ =(1/5) ∫_0 ^(π/2) (sin^2 θ)^3 dθ =(1/5) ∫_0 ^(π/2) {((1−cos(2θ))/2)}^3 dθ =(1/(40)) ∫_0 ^(π/2) {Σ_(k=0) ^3 C_3 ^k (−1)^k (cos(2θ))^k }dθ =(1/(40)) ∫_0 ^(π/2) {1 −3cos(2θ) +3cos^2 (2θ) −cos^3 (2θ)}dθ =(π/(80)) −(3/(40)) ∫_0 ^(π/2) cos(2θ)dθ +(3/(40)) ∫_0 ^(π/2) ((1+cos(4θ))/2)dθ −(1/(40)) ∫_0 ^(π/2) cos(2θ)((1+cos(4θ))/2) dθ =(π/(80)) +(3/(160)) −(1/(80)) ∫_0 ^(π/2) (1+cos(4θ))cos(2θ) ....be continued...](https://www.tinkutara.com/question/Q40205.png)
$${let}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \left(\mathrm{2}−{x}\right)}{dx} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:{x}\sqrt{{x}\left(\mathrm{2}−{x}\right)}\:\:\:\:\:{changement}\:{x}=\mathrm{2}{sin}^{\mathrm{2}} \theta\:{give} \\ $$$${I}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{2}\:{sin}^{\mathrm{2}} \theta\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \theta\left(\mathrm{2}−\mathrm{2}{sin}^{\mathrm{2}} \theta\right)}\:\mathrm{4}\:{sin}\theta\:{cos}\theta{d}\theta \\ $$$$=\mathrm{16}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \theta\:{sin}\theta\:{cos}\theta\:{sin}\theta\:{cos}\theta\:{d}\theta \\ $$$$=\mathrm{16}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}^{\mathrm{4}} \theta\:{cos}^{\mathrm{2}} \theta{d}\theta{d}\theta\:\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\theta\left(\:{cos}\theta\:{sin}^{\mathrm{4}} \theta\right){d}\theta \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{5}}\:{sin}^{\mathrm{5}} \theta\:{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}}{\mathrm{5}}\:{sin}^{\mathrm{5}} \theta\:{sin}\theta\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\left({sin}^{\mathrm{2}} \theta\right)^{\mathrm{3}} \:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left\{\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right\}^{\mathrm{3}} {d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{40}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left\{\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} \:{C}_{\mathrm{3}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} \left({cos}\left(\mathrm{2}\theta\right)\right)^{{k}} \right\}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{40}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\left\{\mathrm{1}\:−\mathrm{3}{cos}\left(\mathrm{2}\theta\right)\:+\mathrm{3}{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)\:−{cos}^{\mathrm{3}} \left(\mathrm{2}\theta\right)\right\}{d}\theta \\ $$$$=\frac{\pi}{\mathrm{80}}\:−\frac{\mathrm{3}}{\mathrm{40}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\left(\mathrm{2}\theta\right){d}\theta\:\:+\frac{\mathrm{3}}{\mathrm{40}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}{d}\theta \\ $$$$−\frac{\mathrm{1}}{\mathrm{40}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}\left(\mathrm{2}\theta\right)\frac{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}\:{d}\theta \\ $$$$=\frac{\pi}{\mathrm{80}}\:+\frac{\mathrm{3}}{\mathrm{160}}\:−\frac{\mathrm{1}}{\mathrm{80}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)\right){cos}\left(\mathrm{2}\theta\right) \\ $$$$….{be}\:{continued}… \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 25/Jul/18
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)\right\}{cos}\left(\mathrm{2}\theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\left(\mathrm{2}\theta\right){d}\theta\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}\theta\right){cos}\left(\mathrm{4}\theta\right){d}\theta \\ $$$$=\mathrm{0}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left\{{cos}\left(\mathrm{6}\theta\right)\:+{cos}\left(\mathrm{2}\theta\right)\right\}{d}\theta\:=\mathrm{0}\:\Rightarrow \\ $$$${I}\:=\mathrm{16}\left\{\:\frac{\mathrm{3}\pi}{\mathrm{160}}\right\}\:\Rightarrow\:{I}\:\:=\frac{\mathrm{3}\pi}{\mathrm{10}}\:. \\ $$