How do you use integration by parts to establish the reduction formula $intcos^n(x) dx = (1/n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx$ ?

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How do you use integration by parts to establish the reduction formula $intcos^n(x) dx = (1/n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx$ ?

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Answer 1 · 2014-09-11T18:29:02

Let
$I=int cos^nxdx$ .

Let $u=cos^{n-1}x$ and $dv=cosxdx$
$Rightarrow du=-(n-1)cos^{n-2}x sinxdx$ and $v=sinx$

By Integration by Parts,
$I=cos^{n-1}x sinx+(n-1)int cos^{n-2}xsin^2xdx$
By $sin^2x=1-cos^2x$ ,
$I=cos^{n-1}x sinx+(n-1)int cos^{n-2}x(1-cos^2x)dx$
By splitting the last integral,
$I=cos^{n-1}x sinx+(n-1)int cos^{n-2}xdx-(n-1)I$
By adding $(n-1)I$ ,
$nI=cos^{n-1}x sinx+(n-1)int cos^{n-2}xdx$
By dividing by $n$ ,
$I=1/ncos^{n-1}xsinx+{n-1}/nintcos^{n-2}xdx$

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How do you use integration by parts to establish the reduction formula $intcos^n(x) dx = (1/n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx$ ?

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How do you use integration by parts to establish the reduction formula intcos^n(x) dx = (1/n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dxintcos^n(x) dx = (1/n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx ?

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How do you use integration by parts to establish the reduction formula $intcos^n(x) dx = (1/n)cos^(n-1)(x)sin(x)+(n-1)/nintcos^(n-2)(x)dx$ ?