How do you evaluate the integral $int sin^3xsqrt(1-cosx)$ ?

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Answer 1 · 2017-01-20T10:29:13

$int sin^3x sqrt (1-cosx)dx = 2/7(1-cosx)^(5/2) (9/5+cosx) + C$

Write:

$sin^3x = sinx * sin^2x = sinx (1-cos^2x) = sinx (1-cosx)(1+cosx)$

so that:

$int sin^3x sqrt (1-cosx)dx = int sinx (1+cosx)(1-cosx)^(3/2) dx$

substitute:

$t = 1-cosx$
$dt = sinx dx$

$int sin^3x sqrt (1-cosx)dx = int t^(3/2)(2-t)dt= 2int t^(3/2)dt - int t^(5/2)dt = 4/5t^(5/2) - 2/7t^(7/2) + C = 2/7t^(5/2) (14/5-t) + C$

Substituting back:

$int sin^3x sqrt (1-cosx)dx = 2/7(1-cosx)^(5/2) (9/5+cosx) + C$

How do you evaluate the integral int sin^3xsqrt(1-cosx)int sin^3xsqrt(1-cosx)?