How do you evaluate the integral $\int {cos}^{3} x {sin}^{3} x$ $int cos^3xsin^3x$ ?

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How do you evaluate the integral $\int {cos}^{3} x {sin}^{3} x$ $int cos^3xsin^3x$ ?

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Answer 1 · 2017-01-04T15:15:17

$= \frac{1}{4} {sin}^{4} x - \frac{1}{6} {sin}^{6} x + C$ $=1/4sin^4x - 1/6sin^6x + C$

Explanation:

We want to rewrite so that either sine or cosine is all by itself with a power of $1$ $1$ and use a u-substitution to eliminate it, leaving all terms in u that are easier to integrate.

$= \int cos x ({cos}^{2} x {sin}^{3} x)$ $=int cosx(cos^2xsin^3x)$

Rewrite ${cos}^{2} x$ $cos^2x$ using the pythagorean identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$ .

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

$= \int cos x ({sin}^{3} x - {sin}^{5} x) d x$ $=int cosx(sin^3x - sin^5x)dx$

The derivative of $sin x$ $sinx$ is $cos x$ $cosx$ . Accordingly, we take the substitution $u = sin x$ $u = sinx$ . Then $d u = cos x d x$ $du = cosxdx$ and $d x = \frac{d u}{cos x}$ $dx = (du)/cosx$

$= \int cos x (u^{3} - u^{5}) \cdot \frac{d u}{cos x}$ $=int cosx(u^3 - u^5) * (du)/cosx$

This will eliminate, leaving us only with u's.

$= \int u^{3} - u^{5} d u$ $=int u^3 - u^5 du$

Integrate using $x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $x^ndx = x^(n + 1)/(n + 1) + C$ .

$= \frac{1}{4} u^{4} - \frac{1}{6} u^{6} + C$ $=1/4u^4 - 1/6u^6 + C$

Resubstitute:

$= \frac{1}{4} {(sin x)}^{4} - \frac{1}{6} {(sin x)}^{6} + C$ $=1/4(sinx)^4 - 1/6(sinx)^6 + C$

$= \frac{1}{4} {sin}^{4} x - \frac{1}{6} {sin}^{6} x + C$ $=1/4sin^4x - 1/6sin^6x + C$

Hopefully this helps!

Answer 2 · 2017-01-04T16:01:27

The answer is $= \frac{1}{192} cos 6 x - \frac{3}{64} cos 2 x + C$ $=1/192cos6x-3/64cos2x+C$

Explanation:

${cos}^{3} x = \frac{1}{4} (cos 3 x + 3 cos x)$ $cos^3x=1/4(cos3x+3cosx)$

${sin}^{3} x = \frac{1}{4} (3 sin x - sin 3 x)$ $sin^3x=1/4(3sinx-sin3x)$

${cos}^{3} x \cdot {sin}^{3} x = \frac{1}{16} (cos 3 x + 3 cos x) (3 sin x - sin 3 x)$ $cos^3x*sin^3x=1/16(cos3x+3cosx)(3sinx-sin3x)$

$= \frac{1}{16} (3 cos 3 x sin x - cos 3 x sin 3 x + 9 cos x sin x - 3 sin 3 x cos x)$ $=1/16(3cos3xsinx-cos3xsin3x+9cosxsinx-3sin3xcosx)$

$9 cos x sin x = \frac{9}{2} sin 2 x$ $9cosxsinx=9/2sin2x$

$cos 3 x sin 3 x = \frac{1}{2} sin 6 x$ $cos3xsin3x=1/2sin6x$

$- 3 sin 3 x cos x + 3 cos 3 x sin x = - 3 (sin (3 x - x)) = - 3 sin 2 x$ $-3sin3xcosx+3cos3xsinx=-3(sin(3x-x))=-3sin2x$

Therefore,

${cos}^{3} x \cdot {sin}^{3} x = \frac{1}{16} (\frac{9}{2} sin 2 x - 3 sin 2 x - \frac{1}{2} sin 6 x)$ $cos^3x*sin^3x=1/16(9/2sin2x-3sin2x-1/2sin6x)$

$= \frac{3}{32} sin 2 x - \frac{1}{32} sin 6 x$ $=3/32sin2x-1/32sin6x$

So,

$\int {cos}^{3} x \cdot {sin}^{3} x d x = \frac{3}{32} \int sin 2 x d x - \frac{1}{32} \int sin 6 x d x$ $intcos^3x*sin^3xdx=3/32intsin2xdx-1/32intsin6xdx$

$= (\frac{3}{32} \cdot - cos 2 x \cdot \frac{1}{2}) - (\frac{1}{32} \cdot - cos 6 x \cdot \frac{1}{6}) + C$ $=(3/32*-cos2x*1/2)-(1/32*-cos6x*1/6)+C$

$= \frac{1}{192} cos 6 x - \frac{3}{64} cos 2 x + C$ $=1/192cos6x-3/64cos2x+C$

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