How do you find the integral of ${cos}^{6} (x)$ $cos^6(x)$ ?

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How do you find the integral of ${cos}^{6} (x)$ $cos^6(x)$ ?

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Answer 1 · 2017-02-18T22:43:48

$\int {cos}^{6} x d x = \frac{sin x {cos}^{5} x}{6} + \frac{sin x {cos}^{3} x}{6} + \frac{2}{5} (sin x cos x) + \frac{2}{5} x + C$ $int cos^6xdx = (sinxcos^5x )/6+ (sinxcos^3x)/6 + 2/5 (sinxcosx) + 2/5x +C$

Explanation:

Integrate by parts:

$\int {cos}^{6} x d x = \int {cos}^{5} x cos x d x = \int {cos}^{5} x d (sin x)$ $int cos^6xdx = int cos^5x cosx dx = int cos^5x d(sinx)$

$\int {cos}^{6} x d x = sin x {cos}^{5} x + 5 \int {cos}^{4} x {sin}^{2} x d x$ $int cos^6xdx = sinxcos^5x + 5 int cos^4x sin^2x dx$

$\int {cos}^{6} x d x = sin x {cos}^{5} x + 5 \int {cos}^{4} x (1 - {cos}^{2} x) d x$ $int cos^6xdx = sinxcos^5x + 5 int cos^4x (1 - cos^2x) dx$

$\int {cos}^{6} x d x = sin x {cos}^{5} x + 5 \int {cos}^{4} x - 5 \int {cos}^{6} x d x$ $int cos^6xdx = sinxcos^5x + 5 int cos^4x - 5int cos^6x dx$

As the integral is on both sides we can solve for it:

$6 \int {cos}^{6} x d x = sin x {cos}^{5} x + 5 \int {cos}^{4} x d x$ $6 int cos^6xdx = sinxcos^5x + 5 int cos^4xdx$

$\int {cos}^{6} x d x = \frac{sin x {cos}^{5} x}{6} + \frac{5}{6} \int {cos}^{4} x d x$ $int cos^6xdx = (sinxcos^5x )/6+ 5/6 int cos^4xdx$

We can use the same method to reduce the degree of $cos x$ $cosx$ again:

$\int {cos}^{4} x d x = \int {cos}^{3} x d (sin x)$ $int cos^4xdx = int cos^3x d(sinx)$

$\int {cos}^{4} x d x = sin x {cos}^{3} x + 4 \int {cos}^{2} x {sin}^{2} x d x$ $int cos^4xdx = sinxcos^3x + 4 int cos^2x sin^2x dx$

$\int {cos}^{4} x d x = sin x {cos}^{3} x + 4 \int {cos}^{2} x d x - 4 \int {cos}^{4} x d x$ $int cos^4xdx = sinxcos^3x + 4 int cos^2xdx -4 int cos^4xdx$

$\int {cos}^{4} x d x = \frac{sin x {cos}^{3} x}{5} + \frac{4}{5} \int {cos}^{2} x d x$ $int cos^4xdx = (sinxcos^3x)/5 + 4/5 int cos^2xdx$

And again:

$\int {cos}^{2} x d x = \int cos x d (sin x)$ $int cos^2xdx = int cosx d(sinx)$

$\int {cos}^{2} x d x = sin x cos x + \int {sin}^{2} x d x$ $int cos^2xdx = sinxcosx + int sin^2x dx$

$\int {cos}^{2} x d x = sin x cos x + \int d x - \int {cos}^{2} x d x$ $int cos^2xdx = sinxcosx + int dx - int cos^2xdx$

$\int {cos}^{2} x d x = \frac{sin x cos x}{2} + \frac{1}{2} x + C$ $int cos^2xdx = (sinxcosx)/2 + 1/2x +C$

Putting it all together we have:

$\int {cos}^{6} x d x = \frac{sin x {cos}^{5} x}{6} + \frac{sin x {cos}^{3} x}{6} + \frac{2}{5} (sin x cos x) + \frac{2}{5} x + C$ $int cos^6xdx = (sinxcos^5x )/6+ (sinxcos^3x)/6 + 2/5 (sinxcosx) + 2/5x +C$

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How do you find the integral of ${cos}^{6} (x)$ $cos^6(x)$ ?

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