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# What is the integral of cos^5(x)?

Dec 16, 2014

$= \sin x + \frac{{\sin}^{5} x}{5} - \frac{2}{3} \cdot {\sin}^{3} x + c$, where $c$ is a constant

Explanation :

$= \int \left({\cos}^{5} x\right) \mathrm{dx}$

From trigonometric identity, which is

${\cos}^{2} x + {\sin}^{2} x = 1$, $\implies {\cos}^{2} x = 1 - \sin 2 x$

$= \int \left({\cos}^{4} x\right) \cdot \cos \left(x\right) \mathrm{dx}$

$= \int {\left({\cos}^{2} x\right)}^{2} \cdot \cos \left(x\right) \mathrm{dx}$

$= \int {\left(1 - {\sin}^{2} x\right)}^{2} \cdot \cos \left(x\right) \mathrm{dx}$ .. $\left(i\right)$

let's assume $\sin x = t$, $\implies \left(\cos x\right) \mathrm{dx} = \mathrm{dt}$

substituting this in the $\left(i\right)$, we get

$= \int {\left(1 - {t}^{2}\right)}^{2} \mathrm{dt}$

Now using expansion of ${\left(1 - y\right)}^{2} = 1 + {y}^{2} - 2 y$, yields,

$= \int \left(1 + {t}^{4} - 2 {t}^{2}\right) \mathrm{dt}$

$= \int \mathrm{dt} + \int {t}^{4} \mathrm{dt} - 2 \int {t}^{2} \mathrm{dt}$

$= t + {t}^{5} / 5 - 2 \cdot {t}^{3} / 3 + c$, where $c$ is a constant

$= t + {t}^{5} / 5 - \frac{2}{3} \cdot {t}^{3} + c$, where $c$ is a constant

now substituting $t$ back gives,

$= \sin x + {\left(\sin x\right)}^{5} / 5 - \frac{2}{3} \cdot {\left(\sin x\right)}^{3} + c$, where $c$ is a constant

$= \sin x + \frac{{\sin}^{5} x}{5} - \frac{2}{3} \cdot {\sin}^{3} x + c$, where $c$ is a constant