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

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Answer 1 · 2014-09-19T20:38:19

Let $I=int sin^nx dx$ .

By pulling out one of $sin x$ 's,

$I=int sinx cdot sin^{n-1}x dx$

Let $u=sin^{n-1}x$ and $dv=sinx dx$
$Rightarrow du=(n-1)sin^{n-2}xcosx$ and $v=-cosx$

by Integration by Parts,

$=-sin^{n-1}xcosx+(n-1)int sin^{n-2}xcos^2xdx$

by the trig identity $cos^2x=1-sin^2x$ ,

$=-sin^{n-1}xcosx+(n-1)int sin^{n-2}x(1-sin^2x)dx$

$=-sin^{n-1}xcosx+(n-1)int sin^{n-2}xdx-(n-1)I$

By adding $(n-1)I$

$Rightarrow nI=-sin^{n-1}xcosx+(n-1)int sin^{n-2}xdx$

By dividing by $n$ ,

$I=-1/nsin^{n-1}xcosx+(n-1)/nint sin^{n-2}xdx$

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