Integration Questions

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

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

Answered by alex041103 last updated on 08/Jul/17

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

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

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

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

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

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

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

Commented by alex041103 last updated on 11/Jul/17

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

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

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

Commented by alex041103 last updated on 13/Jul/17

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

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

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