How do you find the integral of ${cos}^{4} (x) d x$ $cos^4 (x) dx$ ?

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How do you find the integral of ${cos}^{4} (x) d x$ $cos^4 (x) dx$ ?

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Answer 1 · 2017-02-28T18:53:36

$\int {cos}^{4} x d x = \frac{sin x {cos}^{3} x}{4} + \frac{3}{8} (cos x sin x) + \frac{3}{8} x$ $int cos^4xdx = (sinxcos^3x)/4 + 3/8(cosxsinx) + 3/8x$

Explanation:

Write: ${cos}^{4} x = {cos}^{3} x \cdot cos x$ $cos^4x = cos^3x*cosx$ and integrate by parts:

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

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

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

Now use the identity:

${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x = 1-cos^2x$

$\int {cos}^{4} x d x = sin x {cos}^{3} x + 3 \int (1 - {cos}^{2} x) {cos}^{2} x d x$ $int cos^4xdx = sinxcos^3x + 3int (1-cos^2x) cos^2xdx$

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

We have now the same integral on both sides and we can solve for it:

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

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

Using the same process:

$\int {cos}^{2} x d x = \int cos x d (sin x) = cos x sin x + \int {sin}^{2} x d x$ $int cos^2x dx = int cosxd(sinx) = cosxsinx + int sin^2xdx$

$\int {cos}^{2} x d x = \int cos x d (sin x) = cos x sin x + \int (1 - {cos}^{2} x) d x$ $int cos^2x dx = int cosxd(sinx) = cosxsinx + int (1-cos^2x)dx$

$\int {cos}^{2} x d x = \int cos x d (sin x) = cos x sin x + x - \int {cos}^{2} x d x$ $int cos^2x dx = int cosxd(sinx) = cosxsinx + x - int cos^2xdx$

$\int {cos}^{2} x d x = \frac{cos x sin x}{2} + \frac{x}{2}$ $int cos^2x dx = (cosxsinx)/2 + x/2$

Substituting in the expression above:

$\int {cos}^{4} x d x = \frac{sin x {cos}^{3} x}{4} + \frac{3}{8} (cos x sin x) + \frac{3}{8} x$ $int cos^4xdx = (sinxcos^3x)/4 + 3/8(cosxsinx) + 3/8x$

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How do you find the integral of ${cos}^{4} (x) d x$ $cos^4 (x) dx$ ?

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