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

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

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Answer 1 · 2015-10-16T16:53:52

See the explanation for one way to do these.

Explanation:

Even $n$ $n$ For $n = 2 k$ $n=2k$ use
$cos (2 x) = 2 {cos}^{2} x - 1$ $cos(2x) = 2cos^2x-1$ to get the power reduction identity:

${cos}^{2} x = \frac{1}{2} (1 + cos (2 x))$ $cos^2x = 1/2(1+cos(2x))$

So

${({cos}^{2} x)}^{k} = {(\frac{1}{2} (1 + cos (2 x)))}^{k}$ $(cos^2x)^k = (1/2(1+cos(2x)))^k$

Expand the power $k$ $k$ and reduce any

Odd $n$ $n$
For $n = 2 k + 1$ $n=2k+1$ rewrite as

${cos}^{2 k + 1} x = {({cos}^{2} x)}^{k} cos x$ $cos^(2k+1)x = (cos^2x)^k cosx$

$= {(1 - {sin}^{2} x)}^{k} cos x$ $= (1-sin^2x)^k cosx$

Now use $cos (2 x) = 1 - 2 {sin}^{2} x$ $cos(2x) = 1-2sin^2x$ to get the power reduction identity:

${sin}^{2} x = \frac{1}{2} (1 - sin (2 x))$ $sin^2x = 1/2(1-sin(2x))$ .

Us power reduction, expanding the power $k$ $k$ , then power reduction again and repeat as needed to get an integral of the form
$\int (k + a_{n} {sin}_{n}^{b_{n}} u + a_{n - 1} {sin}_{2}^{b_{2}} u + \cdot \cdot \cdot + a_{1} {sin}_{1}^{b_{1}} u) cos u d u$ $int(k+a_nsin^(b_n) u + a_(n-1)sin^(b_2) u + * * * +a_1sin^(b_1)u) cosu du$

Then integrate term by term using substitution.

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

1 Answer

Explanation:

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