Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 17603 by tawa tawa last updated on 08/Jul/17

∫_(  0) ^(  a/2)  x^2 (a^2  − x^2 )^(3/2)  dx

$$\int_{\:\:\mathrm{0}} ^{\:\:\mathrm{a}/\mathrm{2}} \:\mathrm{x}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} \:−\:\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{dx} \\ $$

Answered by alex041103 last updated on 08/Jul/17

I = ∫_(  0) ^(  a/2)  x^2 (a^2  − x^2 )^(3/2)  dx  Let x=asinθ (i.e. θ=sin^(−1) ((x/a)))  ⇒dx = acosθ  We use 1−sin^2 θ=cos^2 θ and    θ=sin^(−1) ((x/a)) to change the limits of  integration:  I = a^6 ∫_(  0) ^(    π/6) sin^2 θcos^4 θ dθ  Then we substitute sin^2 θ=1−cos^2 θ  I = a^6 [∫_( 0) ^(   π/6) cos^4 θ dθ − ∫_(  0) ^(   π/6) cos^6 θ dθ]  Let I_n =∫_(  0) ^(   π/6) cos^n θ dθ  ⇒I=a^6 (I_4 −I_6 )  I_n =∫_(  0) ^(   π/6) cos^(n−1) θcosθ dθ  Then we use integration by parts:  u=cos^(n−1) θ                                     dv=cosθ dθ  du=(n−1)(−sinθ)cos^(n−2) θ    v=sinθ  ∫_a ^b u dv = [uv]_a ^b  − ∫_a ^b v du  I_n  = [cos^(n−1) θsinθ]_0 ^(π/6)  + (n−1)∫_0 ^(π/6) sin^2 θcos^(n−2) θ dθ  We know that sin^2 θ=1−cos^2 θ  ⇒I_n  = (3^((1/2)(n−1)) /2^n ) + (n−1)(∫_0 ^(π/6) cos^(n−2) θ dθ − ∫_0 ^(π/6) cos^n θ dθ)  But   I_n =∫_0 ^(π/6) cos^n θ dθ  I_(n−2) =∫_0 ^(π/6) cos^(n−2) θ dθ  ⇒I_n  = (3^((n−1)/2) /2^n ) + (n−1)I_(n−2)  − (n−1)I_n   i.e. I_n =(3^((n−1)/2) /(n2^n )) + ((n−1)/n)I_(n−2)   We use the formula for I=a^6 (I_(6−2) −I_6 )  and we get  I=a^6 (I_(6−2) −(((3(√3))/2^7 )+(5/6)I_(6−2) ))  =a^6 ((1/6)I_4 −((3(√3))/2^7 ))  Now wd use the formula for I_n  twice  to evaluate I_4 (i.e. expressing I_4  in terms of I_0 ) and we get  I_4 =((3(√3))/2^6 )+((3(√3))/2^5 )+(π/2^4 )  Now we finaly get for I:  I=∫_0 ^(a/2) x^2 (a^2 −x^2 )^(3/2)  dx = ((a^6 π)/(96))

$${I}\:=\:\int_{\:\:\mathrm{0}} ^{\:\:\mathrm{a}/\mathrm{2}} \:\mathrm{x}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} \:−\:\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{dx} \\ $$$${Let}\:{x}={asin}\theta\:\left({i}.{e}.\:\theta={sin}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)\right) \\ $$$$\Rightarrow{dx}\:=\:{acos}\theta \\ $$$${We}\:{use}\:\mathrm{1}−{sin}^{\mathrm{2}} \theta={cos}^{\mathrm{2}} \theta\:{and}\:\: \\ $$$$\theta={sin}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)\:{to}\:{change}\:{the}\:{limits}\:{of} \\ $$$${integration}: \\ $$$${I}\:=\:{a}^{\mathrm{6}} \underset{\:\:\mathrm{0}} {\overset{\:\:\:\:\pi/\mathrm{6}} {\int}}{sin}^{\mathrm{2}} \theta{cos}^{\mathrm{4}} \theta\:{d}\theta \\ $$$${Then}\:{we}\:{substitute}\:{sin}^{\mathrm{2}} \theta=\mathrm{1}−{cos}^{\mathrm{2}} \theta \\ $$$${I}\:=\:{a}^{\mathrm{6}} \left[\underset{\:\mathrm{0}} {\overset{\:\:\:\pi/\mathrm{6}} {\int}}{cos}^{\mathrm{4}} \theta\:{d}\theta\:−\:\underset{\:\:\mathrm{0}} {\overset{\:\:\:\pi/\mathrm{6}} {\int}}{cos}^{\mathrm{6}} \theta\:{d}\theta\right] \\ $$$${Let}\:{I}_{{n}} =\underset{\:\:\mathrm{0}} {\overset{\:\:\:\pi/\mathrm{6}} {\int}}{cos}^{{n}} \theta\:{d}\theta \\ $$$$\Rightarrow{I}={a}^{\mathrm{6}} \left({I}_{\mathrm{4}} −{I}_{\mathrm{6}} \right) \\ $$$${I}_{{n}} =\underset{\:\:\mathrm{0}} {\overset{\:\:\:\pi/\mathrm{6}} {\int}}{cos}^{{n}−\mathrm{1}} \theta{cos}\theta\:{d}\theta \\ $$$${Then}\:{we}\:{use}\:{integration}\:{by}\:{parts}: \\ $$$${u}={cos}^{{n}−\mathrm{1}} \theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{dv}={cos}\theta\:{d}\theta \\ $$$${du}=\left({n}−\mathrm{1}\right)\left(−{sin}\theta\right){cos}^{{n}−\mathrm{2}} \theta\:\:\:\:{v}={sin}\theta \\ $$$$\underset{{a}} {\overset{{b}} {\int}}{u}\:{dv}\:=\:\left[{uv}\right]_{{a}} ^{{b}} \:−\:\underset{{a}} {\overset{{b}} {\int}}{v}\:{du} \\ $$$${I}_{{n}} \:=\:\left[{cos}^{{n}−\mathrm{1}} \theta{sin}\theta\right]_{\mathrm{0}} ^{\pi/\mathrm{6}} \:+\:\left({n}−\mathrm{1}\right)\underset{\mathrm{0}} {\overset{\pi/\mathrm{6}} {\int}}{sin}^{\mathrm{2}} \theta{cos}^{{n}−\mathrm{2}} \theta\:{d}\theta \\ $$$${We}\:{know}\:{that}\:{sin}^{\mathrm{2}} \theta=\mathrm{1}−{cos}^{\mathrm{2}} \theta \\ $$$$\Rightarrow{I}_{{n}} \:=\:\frac{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}\left({n}−\mathrm{1}\right)} }{\mathrm{2}^{{n}} }\:+\:\left({n}−\mathrm{1}\right)\left(\underset{\mathrm{0}} {\overset{\pi/\mathrm{6}} {\int}}{cos}^{{n}−\mathrm{2}} \theta\:{d}\theta\:−\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{6}} {\int}}{cos}^{{n}} \theta\:{d}\theta\right) \\ $$$${But}\: \\ $$$${I}_{{n}} =\underset{\mathrm{0}} {\overset{\pi/\mathrm{6}} {\int}}{cos}^{{n}} \theta\:{d}\theta \\ $$$${I}_{{n}−\mathrm{2}} =\underset{\mathrm{0}} {\overset{\pi/\mathrm{6}} {\int}}{cos}^{{n}−\mathrm{2}} \theta\:{d}\theta \\ $$$$\Rightarrow{I}_{{n}} \:=\:\frac{\mathrm{3}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}^{{n}} }\:+\:\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \:−\:\left({n}−\mathrm{1}\right){I}_{{n}} \\ $$$${i}.{e}.\:{I}_{{n}} =\frac{\mathrm{3}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} }{{n}\mathrm{2}^{{n}} }\:+\:\frac{{n}−\mathrm{1}}{{n}}{I}_{{n}−\mathrm{2}} \\ $$$${We}\:{use}\:{the}\:{formula}\:{for}\:{I}={a}^{\mathrm{6}} \left({I}_{\mathrm{6}−\mathrm{2}} −{I}_{\mathrm{6}} \right) \\ $$$${and}\:{we}\:{get} \\ $$$${I}={a}^{\mathrm{6}} \left({I}_{\mathrm{6}−\mathrm{2}} −\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{7}} }+\frac{\mathrm{5}}{\mathrm{6}}{I}_{\mathrm{6}−\mathrm{2}} \right)\right) \\ $$$$={a}^{\mathrm{6}} \left(\frac{\mathrm{1}}{\mathrm{6}}{I}_{\mathrm{4}} −\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{7}} }\right) \\ $$$${Now}\:{wd}\:{use}\:{the}\:{formula}\:{for}\:{I}_{{n}} \:{twice} \\ $$$${to}\:{evaluate}\:{I}_{\mathrm{4}} \left({i}.{e}.\:{expressing}\:{I}_{\mathrm{4}} \:{in}\:{terms}\:{of}\:{I}_{\mathrm{0}} \right)\:{and}\:{we}\:{get} \\ $$$${I}_{\mathrm{4}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{6}} }+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{5}} }+\frac{\pi}{\mathrm{2}^{\mathrm{4}} } \\ $$$${Now}\:{we}\:{finaly}\:{get}\:{for}\:{I}: \\ $$$${I}=\underset{\mathrm{0}} {\overset{{a}/\mathrm{2}} {\int}}{x}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \:{dx}\:=\:\frac{{a}^{\mathrm{6}} \pi}{\mathrm{96}} \\ $$$$ \\ $$

Commented by tawa tawa last updated on 08/Jul/17

God bless you sir. i really appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Answered by Arnab Maiti last updated on 13/Jul/17

let  I =∫_0 ^( (a/2)) x^2 (a^2 −x^2 )^(3/2) dx  put x=a sinθ  ∴ dx= a cosθ dθ  ∴ I=∫_0 ^( (π/6)) a^3 sin^2 θ{a^2 (1−sin^2 θ)}^(3/2) cosθ dθ  =a^6 ∫_0 ^(π/6) sin^2 θ cos^4 θ dθ  =a^6 (1/8)∫_0 ^( (π/6)) (2sinθ cosθ)^2 2cos^2 θ dθ  =a^6 ×(1/8)∫_0 ^(π/6) sin^2 2θ (1+cos2θ)dθ  =a^6 ×(1/8)[∫_0 ^(π/6) sin^2 2θ dθ+∫_0 ^(π/6) sin^2 2θ cos2θ dθ]  =a^6 /8×[(1/2)∫_0 ^(π/6) (1−cos4θ)dθ+(1/2)∫_0 ^((√3)/2) (sin 2θ)^2 d(sin 2θ)]  =(a^6 /(16))[ θ−((sin 4θ)/4)]_0 ^(π/6) +(a^6 /(16))[(((p)^3 )/3)]_0 ^((√3)/2)    let sin2θ=p  =(a^6 /(16))[{(π/6)−((√3)/8)}+((3(√3))/(3×8))]  =(a^6 /(16))×(π/6)=((πa^6 )/(96))

$$\mathrm{let}\:\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\:\frac{\mathrm{a}}{\mathrm{2}}} \mathrm{x}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{dx} \\ $$$$\mathrm{put}\:\mathrm{x}=\mathrm{a}\:\mathrm{sin}\theta\:\:\therefore\:\mathrm{dx}=\:\mathrm{a}\:\mathrm{cos}\theta\:\mathrm{d}\theta \\ $$$$\therefore\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{6}}} \mathrm{a}^{\mathrm{3}} \mathrm{sin}^{\mathrm{2}} \theta\left\{\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta\right)\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{cos}\theta\:\mathrm{d}\theta \\ $$$$=\mathrm{a}^{\mathrm{6}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{sin}^{\mathrm{2}} \theta\:\mathrm{cos}^{\mathrm{4}} \theta\:\mathrm{d}\theta \\ $$$$=\mathrm{a}^{\mathrm{6}} \frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{6}}} \left(\mathrm{2sin}\theta\:\mathrm{cos}\theta\right)^{\mathrm{2}} \mathrm{2cos}^{\mathrm{2}} \theta\:\mathrm{d}\theta \\ $$$$=\mathrm{a}^{\mathrm{6}} ×\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\:\left(\mathrm{1}+\mathrm{cos2}\theta\right)\mathrm{d}\theta \\ $$$$=\mathrm{a}^{\mathrm{6}} ×\frac{\mathrm{1}}{\mathrm{8}}\left[\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\:\mathrm{d}\theta+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\:\mathrm{cos2}\theta\:\mathrm{d}\theta\right] \\ $$$$=\mathrm{a}^{\mathrm{6}} /\mathrm{8}×\left[\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \left(\mathrm{1}−\mathrm{cos4}\theta\right)\mathrm{d}\theta+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \left(\mathrm{sin}\:\mathrm{2}\theta\right)^{\mathrm{2}} \mathrm{d}\left(\mathrm{sin}\:\mathrm{2}\theta\right)\right] \\ $$$$=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{16}}\left[\:\theta−\frac{\mathrm{sin}\:\mathrm{4}\theta}{\mathrm{4}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} +\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{16}}\left[\frac{\left(\mathrm{p}\right)^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\mathrm{let}\:\mathrm{sin2}\theta=\mathrm{p} \\ $$$$=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{16}}\left[\left\{\frac{\pi}{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\right\}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{3}×\mathrm{8}}\right] \\ $$$$=\frac{\mathrm{a}^{\mathrm{6}} }{\mathrm{16}}×\frac{\pi}{\mathrm{6}}=\frac{\pi\mathrm{a}^{\mathrm{6}} }{\mathrm{96}} \\ $$

Commented by tawa tawa last updated on 11/Jul/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by alex041103 last updated on 11/Jul/17

You have a mistake  x=a sin θ →dx=cos θ dθ  ∴ I=∫_0 ^( (π/6)) a^3 sin^2 θ{a^2 (1−sin^2 θ)}^(3/2) cos θ dθ  I=a^6 ∫_(   0) ^(π/6) sin^2 θ cos^4 θ dθ ≠ a^6 ∫_(   0) ^(π/6) sin^2 θ cos^3 θ dθ

$$\mathrm{You}\:\mathrm{have}\:\mathrm{a}\:\mathrm{mistake} \\ $$$${x}={a}\:\mathrm{sin}\:\theta\:\rightarrow{dx}=\mathrm{cos}\:\theta\:{d}\theta \\ $$$$\therefore\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{6}}} \mathrm{a}^{\mathrm{3}} \mathrm{sin}^{\mathrm{2}} \theta\left\{\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta\right)\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{cos}\:\theta\:\mathrm{d}\theta \\ $$$${I}={a}^{\mathrm{6}} \underset{\:\:\:\mathrm{0}} {\overset{\pi/\mathrm{6}} {\int}}\mathrm{sin}\:^{\mathrm{2}} \theta\:\mathrm{cos}\:^{\mathrm{4}} \theta\:{d}\theta\:\neq\:{a}^{\mathrm{6}} \underset{\:\:\:\mathrm{0}} {\overset{\pi/\mathrm{6}} {\int}}\mathrm{sin}\:^{\mathrm{2}} \theta\:\mathrm{cos}\:^{\mathrm{3}} \theta\:{d}\theta \\ $$$$ \\ $$

Commented by Arnab Maiti last updated on 12/Jul/17

Now it is corrected. Still it is different answer.

$$\mathrm{Now}\:\mathrm{it}\:\mathrm{is}\:\mathrm{corrected}.\:\mathrm{Still}\:\mathrm{it}\:\mathrm{is}\:\mathrm{different}\:\mathrm{answer}. \\ $$

Commented by alex041103 last updated on 13/Jul/17

You now have another mistake  At last lines where you finish the  integration you set sin^2 2θ=((1−cos2θ)/2)  which is not true  ((1−cos2θ)/2)=sin^2 θ≠sin^2 2θ

$${You}\:{now}\:{have}\:{another}\:{mistake} \\ $$$${At}\:{last}\:{lines}\:{where}\:{you}\:{finish}\:{the} \\ $$$${integration}\:{you}\:{set}\:{sin}^{\mathrm{2}} \mathrm{2}\theta=\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}} \\ $$$${which}\:{is}\:{not}\:{true} \\ $$$$\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}}={sin}^{\mathrm{2}} \theta\neq{sin}^{\mathrm{2}} \mathrm{2}\theta \\ $$

Commented by Arnab Maiti last updated on 13/Jul/17

The answer is indentical now.  Thank you!!

$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{indentical}\:\mathrm{now}. \\ $$$$\mathrm{Thank}\:\mathrm{you}!! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com