What is the integral of ${cos}^{5} (x)$ $cos^5(x)$ ?

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Answer 1 · 2014-12-16T04:02:37

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

Explanation :

$= \int ({cos}^{5} x) d x$ $=int(cos^5x) dx$

From trigonometric identity, which is

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x=1$ , $\Rightarrow {cos}^{2} x = 1 - sin 2 x$ $=>cos^2x=1-sin2x$

$= \int ({cos}^{4} x) \cdot cos (x) d x$ $=int(cos^4x)*cos(x) dx$

$= \int {({cos}^{2} x)}^{2} \cdot cos (x) d x$ $=int(cos^2x)^2*cos(x) dx$

$= \int {(1 - {sin}^{2} x)}^{2} \cdot cos (x) d x$ $=int(1-sin^2x)^2*cos(x) dx$ .. $(i)$ $(i)$

let's assume $sin x = t$ $sinx = t$ , $\Rightarrow (cos x) d x = d t$ $=> (cosx) dx= dt$

substituting this in the $(i)$ $(i)$ , we get

$= \int {(1 - t^{2})}^{2} d t$ $=int(1-t^2)^2dt$

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

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

$= \int d t + \int t^{4} d t - 2 \int t^{2} d t$ $=intdt+intt^4dt-2intt^2dt$

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

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

now substituting $t$ $t$ back gives,

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

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

What is the integral of cos5(x)cos^5(x)?