What is $\int {sin}^{2} t {cos}^{4} t d t$ $int sin^(2)t cos^(4)t dt$ ?

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What is $\int {sin}^{2} t {cos}^{4} t d t$ $int sin^(2)t cos^(4)t dt$ ?

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Answer 1 · 2016-03-26T12:46:20

$\int \frac{1}{8} \cdot {(2 sin t cos t)}^{2} \cdot (2 {cos}^{2} t) d t$ $int1/8*(2sintcost)^2*(2cos^2t)dt$
$= \int \frac{1}{8} \cdot {sin}^{2} 2 t \cdot (1 + cos 2 t) d t$ $=int1/8*sin^2 2t*(1+cos2t)dt$
$= \int \frac{1}{16} \cdot 2 {sin}^{2} 2 t \cdot (1 + cos 2 t) d t$ $=int1/16*2sin^2 2t*(1+cos2t)dt$
$= \int \frac{1}{16} \cdot (1 - cos 4 t) \cdot (1 + cos 2 t) d t$ $=int1/16*(1-cos4t)*(1+cos2t)dt$

$= \frac{1}{16} \int (1 - cos 4 t) \cdot (1 + cos 2 t) d t$ $=1/16int(1-cos4t)*(1+cos2t)dt$

$= \frac{1}{16} [\int d t + \int cos 2 t d t - \int cos 4 t d t - \frac{1}{2} \int (2 cos 4 t cos 2 t d t)]$ $=1/16[intdt+intcos2tdt-intcos4tdt-1/2int(2cos4tcos2tdt)]$
$= \frac{1}{16} [\int d t + \int cos 2 t d t - \int cos 4 t d t - \frac{1}{2} \int (cos 6 t + cos 2 t) d t]$ $=1/16[intdt+intcos2tdt-intcos4tdt-1/2int(cos6t+cos2t)dt]$

$= \frac{1}{16} [t + \frac{1}{2} sin 2 t - \frac{1}{4} sin 4 t - \frac{1}{12} sin 6 t - \frac{1}{4} sin 2 t + c]$ $=1/16[t+1/2sin2t-1/4sin4t-1/12sin6t-1/4sin2t+c]$
$= \frac{1}{16} [t + \frac{1}{4} sin 2 t - \frac{1}{4} sin 4 t - \frac{1}{12} sin 6 t + c]$ $=1/16[t+1/4sin2t-1/4sin4t-1/12sin6t+c]$

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What is $\int {sin}^{2} t {cos}^{4} t d t$ $int sin^(2)t cos^(4)t dt$ ?

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