Question

If $\int {\cos }^{5}xdx={}^n{C}_0\cdot \sin x-{}^n{C}_1\cdot \frac{{\sin }^{3}x}{3}+{}^n{C}_2\cdot \frac{{\sin }^{5}x}{5}+C,$ then the value of ${}^{(2n+3)}{C}_2$ is equal to
(where $C$ is integration constant)

Answer 1

$\int {\cos }^{5}xdx=\int {\cos }^{4}x\cdot \cos xdx$
$=\int (1-{\sin }^{2}x{)}^2\cos xdx$
$=\int (1-t^2{)}^{2}dt,$ where $t=\sin x$
$=\int (1-2t^2+t^4)dt=t-\frac{2t^3}{3}+\frac{t^5}{5}+C$
$={}^2{C}_0\cdot \sin x-{}^2{C}_1\cdot \frac{{\sin }^{3}x}{3}+{}^2{C}_2\cdot \frac{{\sin }^{5}x}{5}+C$
So, $n=2$
$\therefore {}^{(2n+3)}{C}_2={}^7{C}_2=21$