Question Number 132121 by abdullahquwatan last updated on 11/Feb/21
$$\int_{\mathrm{2}} ^{\mathrm{8}} \frac{\sqrt{{x}}}{\:\sqrt{\mathrm{10}−{x}}\:+\sqrt{{x}}}\:\mathrm{dx} \\ $$
Commented by Dwaipayan Shikari last updated on 11/Feb/21
$$\mathrm{3} \\ $$
Commented by abdullahquwatan last updated on 11/Feb/21
$${thx} \\ $$$${answer}\:{i}\:{already}\:{know},\:{the}\:{way}\:{don}'{t}\:{know} \\ $$
Answered by EDWIN88 last updated on 11/Feb/21
![let (√x) = (√(10)) sin u or x = 10 sin^2 u⇒dx = 20sin u cos u du I= ∫(((√(10)) sin u)/( (√(10−10sin^2 u)) +(√(10)) sin u))(20 sin u cos u du) I= ∫((20(√(10)) sin^2 u cos u )/( (√(10)) (cos u+sin u))) du I= 10 ∫ ((sin 2u cos u)/(cos 2u))(cos u−sin u)du I=10∫ tan 2u ((1/2)+(1/2)cos 2u)du−(1/2)∫tan 2u sin 2u du =−(5/2)∫((d(cos 2u))/(cos 2u))+5∫sin 2u du −(1/2)∫ ((1−cos^2 2u)/(cos 2u))du =−(5/2)ln ∣cos 2u∣ −(5/2)cos 2u−(1/4)ln ∣sec 2u+tan 2u∣+(1/4)sin 2u I=[−(5/2)ln ∣((5−x)/5)∣−(1/2)(5−x)−(1/4)ln ∣((5+(√(10x−x^2 )))/(5−x))∣+((√(10x−x^2 ))/(20)) ]_2 ^8 I=−(5/2)(ln(3/5)−ln (3/5))−(1/2)(−3−3)−(1/4)(ln 3−ln 3)+((4−4)/(20)) I=−(1/2)(−6)=3](https://www.tinkutara.com/question/Q132130.png)
$$\mathrm{let}\:\sqrt{\mathrm{x}}\:=\:\sqrt{\mathrm{10}}\:\mathrm{sin}\:\mathrm{u}\:\mathrm{or}\:\mathrm{x}\:=\:\mathrm{10}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{u}\Rightarrow\mathrm{dx}\:=\:\mathrm{20sin}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{u}\:\mathrm{du} \\ $$$$\mathrm{I}=\:\int\frac{\sqrt{\mathrm{10}}\:\mathrm{sin}\:\mathrm{u}}{\:\sqrt{\mathrm{10}−\mathrm{10sin}\:^{\mathrm{2}} \mathrm{u}}\:+\sqrt{\mathrm{10}}\:\mathrm{sin}\:\mathrm{u}}\left(\mathrm{20}\:\mathrm{sin}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{u}\:\mathrm{du}\right)\: \\ $$$$\mathrm{I}=\:\int\frac{\mathrm{20}\sqrt{\mathrm{10}}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{u}\:\mathrm{cos}\:\mathrm{u}\:}{\:\sqrt{\mathrm{10}}\:\left(\mathrm{cos}\:\mathrm{u}+\mathrm{sin}\:\mathrm{u}\right)}\:\mathrm{du} \\ $$$$\mathrm{I}=\:\mathrm{10}\:\int\:\frac{\mathrm{sin}\:\mathrm{2u}\:\mathrm{cos}\:\mathrm{u}}{\mathrm{cos}\:\mathrm{2u}}\left(\mathrm{cos}\:\mathrm{u}−\mathrm{sin}\:\mathrm{u}\right)\mathrm{du} \\ $$$$\mathrm{I}=\mathrm{10}\int\:\mathrm{tan}\:\mathrm{2u}\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2u}\right)\mathrm{du}−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{tan}\:\mathrm{2u}\:\mathrm{sin}\:\mathrm{2u}\:\mathrm{du} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mathrm{cos}\:\mathrm{2u}\right)}{\mathrm{cos}\:\mathrm{2u}}+\mathrm{5}\int\mathrm{sin}\:\mathrm{2u}\:\mathrm{du}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{2u}}{\mathrm{cos}\:\mathrm{2u}}\mathrm{du} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{2u}\mid\:−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2u}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{2u}+\mathrm{tan}\:\mathrm{2u}\mid+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2u} \\ $$$$\mathrm{I}=\left[−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\mathrm{5}−\mathrm{x}}{\mathrm{5}}\mid−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}−\mathrm{x}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\frac{\mathrm{5}+\sqrt{\mathrm{10x}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{5}−\mathrm{x}}\mid+\frac{\sqrt{\mathrm{10x}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{20}}\:\right]_{\mathrm{2}} ^{\mathrm{8}} \\ $$$$\mathrm{I}=−\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{ln}\frac{\mathrm{3}}{\mathrm{5}}−\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{5}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{3}−\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{3}\right)+\frac{\mathrm{4}−\mathrm{4}}{\mathrm{20}}\: \\ $$$$\mathrm{I}=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{6}\right)=\mathrm{3} \\ $$
Commented by abdullahquwatan last updated on 11/Feb/21
$${thx} \\ $$
Answered by Dwaipayan Shikari last updated on 11/Feb/21
$$\int_{\mathrm{2}} ^{\mathrm{8}} \frac{\sqrt{{x}}}{\:\sqrt{\mathrm{10}−{x}}+\sqrt{{x}}}{dx}=\int_{\mathrm{2}} ^{\mathrm{8}} \frac{\sqrt{\mathrm{10}−{x}}}{\:\sqrt{{x}}+\sqrt{\mathrm{10}−{x}}}{dx}={I} \\ $$$$\mathrm{2}{I}=\int_{\:\mathrm{2}} ^{\mathrm{8}} \frac{\sqrt{\mathrm{10}−{x}}+\sqrt{{x}}}{\:\sqrt{{x}}+\sqrt{\mathrm{10}−{x}}}{dx} \\ $$$$\mathrm{2}{I}=\left(\mathrm{8}−\mathrm{2}\right)\Rightarrow{I}=\mathrm{3} \\ $$