How do you find the integral of $\int {sin}^{n} (x)$ $int sin^n(x)$ if m or n is an integer?

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How do you find the integral of $\int {sin}^{n} (x)$ $int sin^n(x)$ if m or n is an integer?

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Answer 1 · 2016-10-03T21:26:45

See below.

Explanation:

$\frac{d}{d x} (cos x {sin}^{n - 1} x) = - {sin}^{n} x + (n - 1) {cos}^{2} x {sin}^{n - 2} x =$ $d/(dx)(cos x sin^(n-1)x)=-sin^nx+(n-1)cos^2xsin^(n-2)x =$
$= - {sin}^{n} x + (n - 1) (1 - {sin}^{2} x) {sin}^{n - 2} x =$ $=-sin^n x+(n-1)(1-sin^2x)sin^(n-2)x =$
$= - {sin}^{n} x - (n - 1) {sin}^{n} x + (n - 1) {sin}^{n - 2} x =$ $= -sin^nx-(n-1)sin^n x+(n-1)sin^(n-2)x=$
$= - n {sin}^{n} x + (n - 1) {sin}^{n - 2} x$ $=-nsin^n x+(n-1)sin^(n-2) x$ so calling

$I_{n} = \int {sin}^{n} x d x$ $I_n = int sin^n x dx$ we have

$I_{n} - (\frac{n - 1}{n}) I_{n - 2} = - \frac{1}{n} cos x {sin}^{n - 1} x$ $I_n-((n-1)/n) I_(n-2)=-1/ncos x sin^(n-1)x$ with

$I_{0} = x$ $I_0 = x$ and
$I_{1} = - cos x$ $I_1 = -cos x$

So, with this recurrence equation we can build the response for a given $n$ $n$

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How do you find the integral of $\int {sin}^{n} (x)$ $int sin^n(x)$ if m or n is an integer?

1 Answer

Explanation:

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