Question Number 186840 by cortano12 last updated on 11/Feb/23
$$\:\:\underset{\mathrm{9}} {\overset{\:\mathrm{16}} {\int}}\:\frac{\sqrt{\mathrm{4}−\sqrt{{x}}}}{{x}}\:{dx}\:=? \\ $$
Answered by horsebrand11 last updated on 11/Feb/23
![I=∫_9 ^( 16) ((√(4−(√x)))/x) dx =? let (√x) =4sin^2 t ⇒x=16sin^4 t I=∫ ((√(4−4sin^2 t))/(16sin^4 t)) (64sin^3 t cos t)dt = ∫ ((8cos^2 t dt)/(sin t)) = 8∫ csc t dt −8∫ sin t dt = 8 ln ∣csc t−cot t∣+8cos t + C =8ln ∣((1−cos t)/(sin t))∣+8cos t + C I= 8[ ln ∣((1−cos t)/(sin t))∣+cos t ]_(π/3) ^(π/2) I=8[ 0−(ln ∣(1/( (√3)))∣+(1/2))] I=4ln 3−4](https://www.tinkutara.com/question/Q186850.png)
$$\:\:{I}=\underset{\mathrm{9}} {\overset{\:\mathrm{16}} {\int}}\:\frac{\sqrt{\mathrm{4}−\sqrt{{x}}}}{{x}}\:{dx}\:=? \\ $$$${let}\:\sqrt{{x}}\:=\mathrm{4sin}\:^{\mathrm{2}} {t}\:\Rightarrow{x}=\mathrm{16sin}\:^{\mathrm{4}} {t}\: \\ $$$${I}=\int\:\frac{\sqrt{\mathrm{4}−\mathrm{4sin}\:^{\mathrm{2}} {t}}}{\mathrm{16sin}\:^{\mathrm{4}} {t}}\:\left(\mathrm{64sin}\:^{\mathrm{3}} {t}\:\mathrm{cos}\:{t}\right){dt} \\ $$$$=\:\int\:\frac{\mathrm{8cos}^{\mathrm{2}} \:{t}\:{dt}}{\mathrm{sin}\:{t}}\:=\:\mathrm{8}\int\:\mathrm{csc}\:{t}\:{dt}\:−\mathrm{8}\int\:\mathrm{sin}\:{t}\:{dt} \\ $$$$=\:\mathrm{8}\:\mathrm{ln}\:\mid\mathrm{csc}\:{t}−\mathrm{cot}\:{t}\mid+\mathrm{8cos}\:{t}\:+\:{C} \\ $$$$=\mathrm{8ln}\:\mid\frac{\mathrm{1}−\mathrm{cos}\:{t}}{\mathrm{sin}\:{t}}\mid+\mathrm{8cos}\:{t}\:+\:{C} \\ $$$${I}=\:\mathrm{8}\left[\:\mathrm{ln}\:\mid\frac{\mathrm{1}−\mathrm{cos}\:{t}}{\mathrm{sin}\:{t}}\mid+\mathrm{cos}\:{t}\:\right]_{\pi/\mathrm{3}} ^{\pi/\mathrm{2}} \\ $$$${I}=\mathrm{8}\left[\:\mathrm{0}−\left(\mathrm{ln}\:\mid\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mid+\frac{\mathrm{1}}{\mathrm{2}}\right)\right] \\ $$$${I}=\mathrm{4ln}\:\mathrm{3}−\mathrm{4}\: \\ $$