Question Number 84313 by M±th+et£s last updated on 11/Mar/20
$$\int\:\:\left({a}^{\mathrm{2}/\mathrm{5}} −{x}^{\mathrm{2}/\mathrm{5}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:{dx} \\ $$
Commented by niroj last updated on 11/Mar/20
![∫ (a^(2/5) −x^(2/5) )^(5/2) dx put x= a sinθ dx= a cos θdθ ∫ [ a^(2/5) −a^(2/5) sin^(2/5) θ]^(5/2) a.cosθdθ = ∫ a^((2/5).(5/2)) (1−sin^(2/5) θ)^(5/2) a.cosθdθ = a^2 ∫ cos^((2/5).(5/2)) cosθdθ = a^2 ∫cos^2 θdθ cos^2 θ −1+cos^2 θ=cos2θ 2cos^2 θ=cos2θ+1 = a^2 ∫ (((cos2θ+1)/2))dθ = (a^2 /2)(∫cos2θdθ+∫dθ) = (a^2 /2)( (1/2)sin2θ+θ)+C = (a^2 /4)sin2θ+(a^2 /2)θ +c ∵ x=a sinθ ⇒ θ= sin^(−1) ((x/a)) = (a^2 /4) 2sinθcosθ+(a^2 /2).sin^(−1) ((x/a))+c = (a^2 /2).(x/a)(√(1−sin^2 θ)) + (a^2 /2)sin^(−1) ((x/a))+C = ((ax)/2)(√(1−(x^2 /a^2 ))) + (a^2 /2)sin^(−1) ((x/a))+c = ((ax)/2).((√(a^2 −x^2 ))/a) +(a^2 /2)sin^(−1) ((x/a))+C = ((x(√(a^2 −x^2 )))/2)+(a^2 /2)sin^(−1) ((x/a))+C.](https://www.tinkutara.com/question/Q84314.png)
$$\:\int\:\left(\mathrm{a}^{\mathrm{2}/\mathrm{5}} −\mathrm{x}^{\mathrm{2}/\mathrm{5}} \right)^{\mathrm{5}/\mathrm{2}} \mathrm{dx} \\ $$$$\:\:\mathrm{put}\:\mathrm{x}=\:\mathrm{a}\:\mathrm{sin}\theta \\ $$$$\:\:\:\:\mathrm{dx}=\:\mathrm{a}\:\mathrm{cos}\:\theta\mathrm{d}\theta \\ $$$$\:\:\int\:\left[\:\mathrm{a}^{\mathrm{2}/\mathrm{5}} −\mathrm{a}^{\mathrm{2}/\mathrm{5}} \mathrm{sin}^{\mathrm{2}/\mathrm{5}} \theta\right]^{\frac{\mathrm{5}}{\mathrm{2}}} \mathrm{a}.\mathrm{cos}\theta\mathrm{d}\theta \\ $$$$=\:\int\:\:\mathrm{a}^{\frac{\mathrm{2}}{\mathrm{5}}.\frac{\mathrm{5}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}/\mathrm{5}} \theta\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \mathrm{a}.\mathrm{cos}\theta\mathrm{d}\theta \\ $$$$=\:\mathrm{a}^{\mathrm{2}} \int\:\mathrm{cos}^{\frac{\mathrm{2}}{\mathrm{5}}.\frac{\mathrm{5}}{\mathrm{2}}} \mathrm{cos}\theta\mathrm{d}\theta \\ $$$$=\:\mathrm{a}^{\mathrm{2}} \int\mathrm{cos}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\:\:\:\mathrm{cos}^{\mathrm{2}} \theta\:−\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \theta=\mathrm{cos2}\theta \\ $$$$\:\:\:\:\:\mathrm{2cos}^{\mathrm{2}} \theta=\mathrm{cos2}\theta+\mathrm{1} \\ $$$$\:=\:\mathrm{a}^{\mathrm{2}} \int\:\left(\frac{\mathrm{cos2}\theta+\mathrm{1}}{\mathrm{2}}\right)\mathrm{d}\theta \\ $$$$=\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\left(\int\mathrm{cos2}\theta\mathrm{d}\theta+\int\mathrm{d}\theta\right) \\ $$$$=\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}\theta+\theta\right)+\mathrm{C} \\ $$$$=\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\mathrm{sin2}\theta+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\theta\:+\mathrm{c} \\ $$$$\:\because\:\mathrm{x}=\mathrm{a}\:\mathrm{sin}\theta\:\Rightarrow\:\theta=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{a}}\right) \\ $$$$\:=\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\:\mathrm{2sin}\theta\mathrm{cos}\theta+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}.\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c} \\ $$$$=\:\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{x}}{\mathrm{a}}\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}\:\:\:+\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C} \\ $$$$=\:\:\frac{\mathrm{ax}}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }}\:\:+\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c} \\ $$$$=\:\:\frac{\mathrm{ax}}{\mathrm{2}}.\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }}{\mathrm{a}}\:+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C} \\ $$$$=\:\:\frac{\mathrm{x}\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }}{\mathrm{2}}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}. \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 11/Mar/20
$${thanks}\:{sir} \\ $$
Commented by niroj last updated on 11/Mar/20
$${you}\:{must}\:{welcome}. \\ $$
Answered by TANMAY PANACEA last updated on 11/Mar/20
![x^(1/5) =a^(1/5) sinα x=asin^5 α→dx=a×5sin^4 α×cosα ∫(a^(2/5) −a^(2/5) sin^2 α)^(5/2) ×a×5sin^4 αcosα dα 5∫acos^5 α×asin^4 α×cosαdα 5a^2 ∫(((1+co2α)/2))^3 ×(((1−cos2α)/2))^2 dα ((5a^2 )/2^5 )∫(1+cos2α)(1−cos^2 2α)^2 dα ((5a^2 )/(32))∫(sin^4 2α+sin^4 2αcos2α)dα ((5a^2 )/(32))∫sin^4 2αdα+((5a^2 )/(64))∫sin^4 2α×d(sin2α) I_2 =((5a^2 )/(64))×((sin^5 2α)/5)=(1/(64))sin^5 2α+C_1 I_1 =((5a^2 )/(32))∫[(((1+cos4α)/2))^2 ] dα =((5a^2 )/(128))∫[1+2cos4α+(((1+cos8α)/2))] dα =((5a^2 )/(128)){α+((2sin4α)/4)+(1/2)(α+((sin8α)/8))} where sinα=((x/a))^(1/5) pls check upto this](https://www.tinkutara.com/question/Q84318.png)
$${x}^{\frac{\mathrm{1}}{\mathrm{5}}} ={a}^{\frac{\mathrm{1}}{\mathrm{5}}} {sin}\alpha \\ $$$${x}={asin}^{\mathrm{5}} \alpha\rightarrow{dx}={a}×\mathrm{5}{sin}^{\mathrm{4}} \alpha×{cos}\alpha \\ $$$$\int\left({a}^{\frac{\mathrm{2}}{\mathrm{5}}} −{a}^{\frac{\mathrm{2}}{\mathrm{5}}} {sin}^{\mathrm{2}} \alpha\right)^{\frac{\mathrm{5}}{\mathrm{2}}} ×{a}×\mathrm{5}{sin}^{\mathrm{4}} \alpha{cos}\alpha\:{d}\alpha \\ $$$$\mathrm{5}\int{acos}^{\mathrm{5}} \alpha×{asin}^{\mathrm{4}} \alpha×{cos}\alpha{d}\alpha \\ $$$$\mathrm{5}{a}^{\mathrm{2}} \int\left(\frac{\mathrm{1}+{co}\mathrm{2}\alpha}{\mathrm{2}}\right)^{\mathrm{3}} ×\left(\frac{\mathrm{1}−{cos}\mathrm{2}\alpha}{\mathrm{2}}\right)^{\mathrm{2}} {d}\alpha \\ $$$$\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{5}} }\int\left(\mathrm{1}+{cos}\mathrm{2}\alpha\right)\left(\mathrm{1}−{cos}^{\mathrm{2}} \mathrm{2}\alpha\right)^{\mathrm{2}} {d}\alpha \\ $$$$\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{32}}\int\left({sin}^{\mathrm{4}} \mathrm{2}\alpha+{sin}^{\mathrm{4}} \mathrm{2}\alpha{cos}\mathrm{2}\alpha\right){d}\alpha \\ $$$$\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{32}}\int{sin}^{\mathrm{4}} \mathrm{2}\alpha{d}\alpha+\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{64}}\int{sin}^{\mathrm{4}} \mathrm{2}\alpha×{d}\left({sin}\mathrm{2}\alpha\right) \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{64}}×\frac{{sin}^{\mathrm{5}} \mathrm{2}\alpha}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{64}}{sin}^{\mathrm{5}} \mathrm{2}\alpha+{C}_{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{32}}\int\left[\left(\frac{\mathrm{1}+{cos}\mathrm{4}\alpha}{\mathrm{2}}\right)^{\mathrm{2}} \:\:\right]\:{d}\alpha \\ $$$$=\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{128}}\int\left[\mathrm{1}+\mathrm{2}{cos}\mathrm{4}\alpha+\left(\frac{\mathrm{1}+{cos}\mathrm{8}\alpha}{\mathrm{2}}\right)\right]\:\:{d}\alpha \\ $$$$=\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{128}}\left\{\alpha+\frac{\mathrm{2}{sin}\mathrm{4}\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\alpha+\frac{{sin}\mathrm{8}\alpha}{\mathrm{8}}\right)\right\} \\ $$$${where}\:{sin}\alpha=\left(\frac{{x}}{{a}}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$${pls}\:{check}\:{upto}\:{this} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 11/Mar/20
$${thank}\:{you}\:{sir}\: \\ $$