Januar 3, 2020

von Danyette

Wie finden Sie das Integral von ${sin}^{3} (x) {cos}^{2} (x) d x$ $sin ^ 3 (x) cos ^ 2 (x) dx$ ?

Antworten:

$I = \frac{1}{5} {cos}^{5} x - \frac{1}{3} {cos}^{3} x + C$ $I = 1/5cos^5x-1/3cos^3x+C$

Erläuterung:

$I = \int {sin}^{3} x {cos}^{2} x d x = \int {sin}^{2} x {cos}^{2} x sin x d x$ $I = int sin^3xcos^2xdx = int sin^2xcos^2xsinxdx$
$I = \int (1 - {cos}^{2} x) {cos}^{2} x sin x d x$ $I = int (1-cos^2x)cos^2xsinxdx$
$cos x = t \Rightarrow - sin x d x = d t \Rightarrow sin x d x = - d t$ $cosx=t => -sinxdx=dt => sinxdx=-dt$

$I = \int (1 - t^{2}) t^{2} (- d t) = \int (t^{4} - t^{2}) d t = \frac{t^{5}}{5} - \frac{t^{3}}{3} + C$ $I = int (1-t^2)t^2(-dt) = int (t^4-t^2)dt = t^5/5-t^3/3+C$

$I = \frac{1}{5} {cos}^{5} x - \frac{1}{3} {cos}^{3} x + C$ $I = 1/5cos^5x-1/3cos^3x+C$