Question

Solve :

${\int }_0^{\pi }{\sin }^{9}x{\cos }^{5}xdx$

• A. $\frac{13\cdot \mathrm{11.....1}}{14\cdot \mathrm{12.....2}}\times \frac{\pi }{2}$
• B. $\frac{[9\cdot 7\cdot 5\cdot 3\cdot 1][4\cdot 2]}{14\cdot 12\cdot \mathrm{10.....4}\cdot 2}$
• C. $0$
• D. None of these

Answer 1

Answer: (C)

$0$

Let $I={\int }_0^{\pi }{\sin }^{9}x{\cos }^{5}xdx$

$={\int }_0^{\pi }{\sin }^9(\pi -x){\cos }^5(\pi -x)dx$ $\left[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right]$

$=-{\int }_0^{\pi }{\sin }^{9}x{\cos }^{5}xdx=-I$

$\Rightarrow 2I=0\Rightarrow I=0$