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Question Number 64320 by aliesam last updated on 16/Jul/19

without beta function ∫cos^3 t sin^2 t dt

$${without}\:{beta}\:{function} \\ $$$$\int{cos}^{\mathrm{3}} {t}\:{sin}^{\mathrm{2}} {t}\:{dt}\: \\ $$

Commented by mathmax by abdo last updated on 16/Jul/19

I =∫ sint (sint cos^3 t)dt by parts u=sint and v^′ =sint cos^3 t ⇒I =−(1/4)sint cos^4 t +(1/4)∫ cos^5 t dt +c ∫ cos^5 t dt =∫ cost(cos^2 t)^2 dt =∫ cost (((1+cos(2t))/2))^2 dt =∫ (1/4)cost(1+2cost +cos^2 (2t))dt =(1/4) ∫ cost dt +(1/2) ∫ cos^2 t dt +(1/4) ∫ cost cos^2 (2t)dt =(1/4)sint +(1/4) ∫(1+cos(2t))dt +(1/8) ∫ cost(1+cos(4t))dt =(1/4)sint +(t/4) +(1/8)sin(2t) +(1/8)sint +(1/8) ∫ cost cos(4t)dt =(3/8)sint +(t/4) +(1/8)sin(2t) +(1/8) ∫ cost cos(4t)dt =(3/8)sint +(t/4) +(1/8)sin(2t) +(1/(16))∫(cos(5t)+cos(3t))dt =(3/8)sint +(t/4) +(1/8)sin(2t)+(1/(80))sin(5t)+(1/(48)) sin(3t) ⇒ I =−(1/4)sint cos^4 t +(3/(32))sint +(t/(16)) +(1/(24))sin(2t)+(1/(320))sin(5t)+(1/(192))sin(3t) +c

$${I}\:=\int\:{sint}\:\left({sint}\:{cos}^{\mathrm{3}} {t}\right){dt}\:{by}\:{parts}\:{u}={sint}\:{and}\:{v}^{'} ={sint}\:{cos}^{\mathrm{3}} {t} \\ $$$$\Rightarrow{I}\:=−\frac{\mathrm{1}}{\mathrm{4}}{sint}\:{cos}^{\mathrm{4}} {t}\:+\frac{\mathrm{1}}{\mathrm{4}}\int\:\:{cos}^{\mathrm{5}} {t}\:{dt}\:+{c} \\ $$$$\int\:{cos}^{\mathrm{5}} {t}\:{dt}\:=\int\:{cost}\left({cos}^{\mathrm{2}} {t}\right)^{\mathrm{2}} {dt}\:=\int\:{cost}\:\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} {dt} \\ $$$$=\int\:\frac{\mathrm{1}}{\mathrm{4}}{cost}\left(\mathrm{1}+\mathrm{2}{cost}\:+{cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int\:{cost}\:{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:{cos}^{\mathrm{2}} {t}\:{dt}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int\:{cost}\:{cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sint}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int\left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{cost}\left(\mathrm{1}+{cos}\left(\mathrm{4}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sint}\:+\frac{{t}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{2}{t}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}{sint}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{cost}\:{cos}\left(\mathrm{4}{t}\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}{sint}\:+\frac{{t}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{2}{t}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{cost}\:{cos}\left(\mathrm{4}{t}\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}{sint}\:+\frac{{t}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{2}{t}\right)\:+\frac{\mathrm{1}}{\mathrm{16}}\int\left({cos}\left(\mathrm{5}{t}\right)+{cos}\left(\mathrm{3}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}{sint}\:+\frac{{t}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{2}{t}\right)+\frac{\mathrm{1}}{\mathrm{80}}{sin}\left(\mathrm{5}{t}\right)+\frac{\mathrm{1}}{\mathrm{48}}\:{sin}\left(\mathrm{3}{t}\right)\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{4}}{sint}\:{cos}^{\mathrm{4}} {t}\:+\frac{\mathrm{3}}{\mathrm{32}}{sint}\:+\frac{{t}}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{24}}{sin}\left(\mathrm{2}{t}\right)+\frac{\mathrm{1}}{\mathrm{320}}{sin}\left(\mathrm{5}{t}\right)+\frac{\mathrm{1}}{\mathrm{192}}{sin}\left(\mathrm{3}{t}\right)\:+{c} \\ $$$$ \\ $$$$ \\ $$

Answered by Tanmay chaudhury last updated on 16/Jul/19

∫(1−sin^2 t)sin^2 tcostdt ∫(a^2 −a^4 )da a=sint da=costdt (a^3 /3)−(a^5 /5)+c ((sin^3 t)/3)−((sin^5 t)/5)+c

$$\int\left(\mathrm{1}−{sin}^{\mathrm{2}} {t}\right){sin}^{\mathrm{2}} {tcostdt} \\ $$$$\int\left({a}^{\mathrm{2}} −{a}^{\mathrm{4}} \right){da}\:\:\:{a}={sint}\:\:\:{da}={costdt} \\ $$$$\frac{{a}^{\mathrm{3}} }{\mathrm{3}}−\frac{{a}^{\mathrm{5}} }{\mathrm{5}}+{c} \\ $$$$\frac{{sin}^{\mathrm{3}} {t}}{\mathrm{3}}−\frac{{sin}^{\mathrm{5}} {t}}{\mathrm{5}}+{c} \\ $$

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