How do you find the integral of $({cos}^{4} x sin x) d x$ $(cos^4 x sin x) dx$ from $[0, \frac{π}{3}]$ $[0, pi/3]$ ?

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How do you find the integral of $({cos}^{4} x sin x) d x$ $(cos^4 x sin x) dx$ from $[0, \frac{π}{3}]$ $[0, pi/3]$ ?

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Answer 1 · 2015-04-17T03:50:31

Have a look:
$\int {cos}^{4} (x) sin (x) d x =$ $intcos^4(x)sin(x)dx=$
but: $d [cos (x)] = - sin (x) d x$ $d[cos(x)]=-sin(x)dx$
$= - \int {cos}^{4} (x) d [cos (x)] =$ $=-intcos^4(x)d[cos(x)]=$
$= - \frac{{cos}^{5} (x)}{5} {∣ ∣ ∣}_{0}^{\frac{π}{3}}$ $=-cos^5(x)/5|_0^(pi/3)$ $= \frac{31}{160}$ $=31/160$

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How do you find the integral of $({cos}^{4} x sin x) d x$ $(cos^4 x sin x) dx$ from $[0, \frac{π}{3}]$ $[0, pi/3]$ ?

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How do you find the integral of $({cos}^{4} x sin x) d x$ $(cos^4 x sin x) dx$ from $[0, \frac{π}{3}]$ $[0, pi/3]$ ?