How do you integrate $((cosx)^3 sinx )/(4sqrt{1-(cosx)^4})dx$ ?

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Answer 1 · 2015-04-17T19:01:22

This integral is very easy, explanation :

$1/4int(cos^3(x)*sin(x))/(sqrt(1-cos^4(x)))dx$

Let's $t=cos(x)$

So $dt = -sin(x)dx$

$1/2intt^3/(2(sqrt(1-t^4)))d(-t) = 1/2int(-t^3)/(2sqrt(1-t^4))dt$

Let's $u = 1-t^4$

So $du = -4t^3dt$

$=>1/8int(-4t^3)/(2sqrt(1-t^4))dt$

$=>1/8int1/(2sqrt(u))du$

$= 1/8[sqrt(u)]$

Substitute back for $u = 1-t^4$ and $t=cos(x)$

$= 1/8[sqrt(1-cos^4(x))] + C$

With habits you can directly do $t = 1 - cos^4(x)$

$dt = 4sin(x)cos^3(x)$

How do you integrate ((cosx)^3 sinx )/(4sqrt{1-(cosx)^4})dx((cosx)^3 sinx )/(4sqrt{1-(cosx)^4})dx?