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5cos-5-tsin-t-




Question Number 21591 by Isse last updated on 29/Sep/17
5cos^5 tsin t
$$\mathrm{5cos}\:^{\mathrm{5}} {t}\mathrm{sin}\:{t} \\ $$
Commented by mrW1 last updated on 29/Sep/17
what′s your concrete question?
$$\mathrm{what}'\mathrm{s}\:\mathrm{your}\:\mathrm{concrete}\:\mathrm{question}? \\ $$
Answered by $@ty@m last updated on 29/Sep/17
Letcost=u   ⇒−sintdt=du   ∫5cos^5 tsin tdt  =∫5u^5 (−du)  =−5∫u^5 du  =−5(u^6 /6)+C  =((−5)/6)cos^6 t +C
$${Let}\mathrm{cos}{t}={u}\: \\ $$$$\Rightarrow−\mathrm{sin}{tdt}={du}\: \\ $$$$\int\mathrm{5cos}\:^{\mathrm{5}} {t}\mathrm{sin}\:{tdt} \\ $$$$=\int\mathrm{5}{u}^{\mathrm{5}} \left(−{du}\right) \\ $$$$=−\mathrm{5}\int{u}^{\mathrm{5}} {du} \\ $$$$=−\mathrm{5}\frac{{u}^{\mathrm{6}} }{\mathrm{6}}+\mathrm{C} \\ $$$$=\frac{−\mathrm{5}}{\mathrm{6}}\mathrm{cos}^{\mathrm{6}} {t}\:+\mathrm{C} \\ $$

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