Question Number 178832 by Spillover last updated on 22/Oct/22
Answered by ARUNG_Brandon_MBU last updated on 22/Oct/22
$${a}.\:\int\frac{\mathrm{y}\left(\mathrm{y}−\mathrm{8}\right)}{\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} }{d}\mathrm{y}=\int\frac{\left(\left(\mathrm{y}−\mathrm{4}\right)+\mathrm{4}\right)\left(\left(\mathrm{y}−\mathrm{4}\right)−\mathrm{4}\right)}{\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} }{d}\mathrm{y} \\ $$$$\:\:\:\:=\int\frac{\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{16}}{\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} }{d}\mathrm{y}=\int\left(\mathrm{1}−\frac{\mathrm{16}}{\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} }\right){d}\mathrm{y} \\ $$$$\:\:\:\:=\mathrm{y}+\frac{\mathrm{16}}{\mathrm{y}−\mathrm{4}}+{C}=\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{4y}+\mathrm{16}}{\mathrm{y}−\mathrm{4}}+{C} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 22/Oct/22
![b. ∫_1 ^(10) xlog_(10) xdx=(1/(ln10))∫_1 ^(10) xlnxdx =(1/(ln10))([(x^2 /2)lnx]_1 ^(10) −(1/2)∫_1 ^(10) xdx) =(1/(ln10))(50ln10−(1/4)[x^2 ]_1 ^(10) ) =(1/(ln10))(50ln10−((99)/4))=50−((99)/(4ln10))](https://www.tinkutara.com/question/Q178841.png)
$${b}.\:\int_{\mathrm{1}} ^{\mathrm{10}} {x}\mathrm{log}_{\mathrm{10}} {xdx}=\frac{\mathrm{1}}{\mathrm{ln10}}\int_{\mathrm{1}} ^{\mathrm{10}} {x}\mathrm{ln}{xdx} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{ln10}}\left(\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}{x}\right]_{\mathrm{1}} ^{\mathrm{10}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{10}} {xdx}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{ln10}}\left(\mathrm{50ln10}−\frac{\mathrm{1}}{\mathrm{4}}\left[{x}^{\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{10}} \right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{ln10}}\left(\mathrm{50ln10}−\frac{\mathrm{99}}{\mathrm{4}}\right)=\mathrm{50}−\frac{\mathrm{99}}{\mathrm{4ln10}} \\ $$
Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22
![c. ∫_a ^b ((xdx)/( (√((x−a)(b−x))))), x=acos^2 θ+bsin^2 θ =a+(b−a)sin^2 θ=b+(a−b)cos^2 θ ⇒dx=(b−a)sin2θdθ =∫_0 ^(π/2) (((acos^2 θ+bsin^2 θ)(b−a)sin2θ)/( (√((b−a)sin^2 θ(b−a)cos^2 θ))))dθ =((b−a)/(∣b−a∣))∫_0 ^(π/2) 2(acos^2 θ+bsin^2 θ)dθ =∫_0 ^(π/2) (a+acos2θ+b−bcos2θ)dθ =[aθ+((asin2θ)/2)+bθ−((bsin2θ)/2)]_0 ^(π/2) =(π/2)(a+b)](https://www.tinkutara.com/question/Q178855.png)
$${c}. \\ $$$$\int_{{a}} ^{{b}} \frac{{xdx}}{\:\sqrt{\left({x}−{a}\right)\left({b}−{x}\right)}},\:{x}={a}\mathrm{cos}^{\mathrm{2}} \theta+{b}\mathrm{sin}^{\mathrm{2}} \theta \\ $$$$={a}+\left({b}−{a}\right)\mathrm{sin}^{\mathrm{2}} \theta={b}+\left({a}−{b}\right)\mathrm{cos}^{\mathrm{2}} \theta \\ $$$$\Rightarrow{dx}=\left({b}−{a}\right)\mathrm{sin2}\theta{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left({a}\mathrm{cos}^{\mathrm{2}} \theta+{b}\mathrm{sin}^{\mathrm{2}} \theta\right)\left({b}−{a}\right)\mathrm{sin2}\theta}{\:\sqrt{\left({b}−{a}\right)\mathrm{sin}^{\mathrm{2}} \theta\left({b}−{a}\right)\mathrm{cos}^{\mathrm{2}} \theta}}{d}\theta \\ $$$$=\frac{{b}−{a}}{\mid{b}−{a}\mid}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}\left({a}\mathrm{cos}^{\mathrm{2}} \theta+{b}\mathrm{sin}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({a}+{a}\mathrm{cos2}\theta+{b}−{b}\mathrm{cos2}\theta\right){d}\theta \\ $$$$=\left[{a}\theta+\frac{{a}\mathrm{sin2}\theta}{\mathrm{2}}+{b}\theta−\frac{{b}\mathrm{sin2}\theta}{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}}\left({a}+{b}\right) \\ $$
Commented by Spillover last updated on 22/Oct/22
$$\mathrm{nice}.\mathrm{thanks} \\ $$
Answered by cortano1 last updated on 22/Oct/22
![(c) x=acos^2 θ+bsin^2 θ =a+(b−a)sin^2 θ { ((x=b⇒θ=(π/2))),((x=a⇒θ=0)) :}⇒dx=(b−a)sin 2θ dθ I=∫_0 ^(π/2) (((a+(b−a)sin^2 θ))/( (√(((b−a)sin^2 θ)((b−a)−(b−a)sin^2 θ))))) (b−a)sin 2θ dθ =∫_0 ^(π/2) ((2(a+(b−a)sin^2 θ))/((b−a)(√(sin^2 2θ )))) (b−a)sin 2θ dθ =∫_0 ^(π/2) (2a+2(b−a)(((1−cos 2θ)/2)))dθ = πa+(b−a)∫_0 ^(π/2) (1−cos 2θ)dθ =πa+(b−a)(π/2)−(1/2)(b−a)[sin 2θ ]_0 ^(π/2) =(a+(1/2)b−(1/2)a)π=(1/2)π(b+a)](https://www.tinkutara.com/question/Q178844.png)
$$\left(\mathrm{c}\right)\:\mathrm{x}=\mathrm{acos}\:^{\mathrm{2}} \theta+\mathrm{bsin}\:^{\mathrm{2}} \theta\:=\mathrm{a}+\left(\mathrm{b}−\mathrm{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\:\begin{cases}{\mathrm{x}=\mathrm{b}\Rightarrow\theta=\frac{\pi}{\mathrm{2}}}\\{\mathrm{x}=\mathrm{a}\Rightarrow\theta=\mathrm{0}}\end{cases}\Rightarrow\mathrm{dx}=\left(\mathrm{b}−\mathrm{a}\right)\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{d}\theta \\ $$$$\mathrm{I}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\left(\mathrm{a}+\left(\mathrm{b}−\mathrm{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta\right)}{\:\sqrt{\left(\left(\mathrm{b}−\mathrm{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta\right)\left(\left(\mathrm{b}−\mathrm{a}\right)−\left(\mathrm{b}−\mathrm{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta\right)}}\:\left(\mathrm{b}−\mathrm{a}\right)\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{d}\theta \\ $$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{2}\left(\mathrm{a}+\left(\mathrm{b}−\mathrm{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta\right)}{\left(\mathrm{b}−\mathrm{a}\right)\sqrt{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta\:}}\:\left(\mathrm{b}−\mathrm{a}\right)\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{d}\theta \\ $$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\left(\mathrm{2a}+\mathrm{2}\left(\mathrm{b}−\mathrm{a}\right)\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}}\right)\right)\mathrm{d}\theta \\ $$$$=\:\pi\mathrm{a}+\left(\mathrm{b}−\mathrm{a}\right)\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)\mathrm{d}\theta \\ $$$$=\pi\mathrm{a}+\left(\mathrm{b}−\mathrm{a}\right)\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{b}−\mathrm{a}\right)\left[\mathrm{sin}\:\mathrm{2}\theta\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{b}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}\right)\pi=\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{b}+\mathrm{a}\right) \\ $$$$ \\ $$
Commented by Spillover last updated on 22/Oct/22
$$\mathrm{nice}.\mathrm{thanks} \\ $$
Answered by Spillover last updated on 22/Oct/22
