What is the integral of ${cos}^{3} x$ $cos^3 x$ ?

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Answer 1 · 2016-05-31T18:25:51

$\int {cos}^{3} x d x = sin x - \frac{{sin}^{3} x}{3} + constant$ $intcos^3x dx = sinx -sin^3x/3 + "constant"$

There are multiple ways to get at this integral which give alternate forms as output. A simple way to do it is to break up the integrand using

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

Which gives us

$\int {cos}^{3} x d x = \int (1 - {sin}^{2} x) cos x d x$ $intcos^3x dx = int(1-sin^2x)cosxdx$

$= \int cos x d x - \int {sin}^{2} x cos x d x$ $= int cosxdx- int sin^2xcosxdx$

We can now do each term separately, where the first is simply:

$\int cos x d x = sin x$ $int cosxdx = sinx$

and the second term can be simplified by making the substitution:

$u = sin x \Rightarrow d u = cos x d x$ $u = sinx implies du = cosxdx$

$\int u^{2} d u = \frac{u^{3}}{3} = \frac{{sin}^{3} x}{3}$ $int u^2du = (u^3)/3 = sin^3x/3$

Putting these together we get

$\int {cos}^{3} x d x = sin x - \frac{{sin}^{3} x}{3} + constant$ $intcos^3x dx = sinx -sin^3x/3 + "constant"$

What is the integral of cos3xcos^3 x?