Devan – Die Kluge Eule

Was ist das Integral von $\int {sin}^{2} (x) . {cos}^{2} (x) d x$ $int sin ^ 2 (x) .cos ^ 2 (x) dx$ ?

Januar 27, 2020

von Devan

Was ist das Integral von $\int {sin}^{2} (x) . {cos}^{2} (x) d x$ $int sin ^ 2 (x) .cos ^ 2 (x) dx$ ? Antworten: $\int {sin}^{2} x {cos}^{2} x d x = \frac{x}{8} - \frac{sin 4 x}{32} + c$ $intsin^2xcos^2xdx=x/8-(sin4x)/32+c$ Erläuterung: As $sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$ $\int {sin}^{2} x {cos}^{2} x d x = \frac{1}{4} \int (4 {sin}^{2} x {cos}^{2} x) d x$ $intsin^2xcos^2xdx=1/4int(4sin^2xcos^2x)dx$ = $\frac{1}{4} \int {sin}^{2} (2 x) d x$ $1/4intsin^2(2x)dx$ = $\frac{1}{4} \int \frac{1 - cos 4 x}{2} d x$ $1/4int(1-cos4x)/2dx$ = $\frac{x}{8} - \frac{1}{8} \int cos 4 x d x$ $x/8-1/8intcos4xdx$ = $\frac{x}{8} - \frac{1}{8} \times \frac{sin 4 x}{4} + c$ $x/8-1/8xx(sin4x)/4+c$ = $\frac{x}{8} - \frac{sin 4 x}{32} + c$ $x/8-(sin4x)/32+c$