What is $int_(pi/8)^((11pi)/12) sin^3x-cosx*cotxdx$ ?

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Answer 1 · 2018-05-13T15:19:02

The answer is $=-0.426$

First, calculate the indefinite integral

$I=int(sin^3x-cosxcotx)dx$

$=intsin^3xdx-intcoscotxdx$

$=I_1-I_2$

$I_1=intsin^3xdx=intsin^2xsinxdx$

$=int(1-cos^2x)sinxdx$

Let $u=cosx$ , $=>$ , $du=-sinxdx$

$I_1=int-(1-u^2)du$

$=u^3/3-u$

$=1/3cos^3x-cosx$

$I_2=intcoscotxdx$

$=int(cos^2xdx)/(sinx)$

$=int((1-sin^2x)dx)/(sinx)$

$=intcscxdx-intsinxdx$

$=-ln(cscx+cotx)+cosx$

Finally,

$I=1/3cos^3x-2cosx+ln(|cscx+cotx|)+C$

The definite integral is

$int_(1/8pi) ^(11/12pi)(sin^3x-cosxcotx)dx$

$=[1/3cos^3x-2cosx+ln(|cscx+cotx|)]_(1/8pi)^(11/12pi)$

$=(1/3cos^3(11/12pi)-2cos(11/12pi)+ln(|csc(11/12pi)+cot(11/12pi)|))-(1/3cos^3(1/8pi)-2cos(1/8pi)+ln(|csc(1/8pi)+cot(1/8pi)|))$

$=-0.426$

What is int_(pi/8)^((11pi)/12) sin^3x-cosx*cotxdxint_(pi/8)^((11pi)/12) sin^3x-cosx*cotxdx?