Question

If $I_n={\int }_0^{\infty }{e}^{-x}(sinx{)}^{n}dx;(n>1)$ then the value of $\frac{101I_{10}}{10I_8}$ is ?

• A. 9
• B. 10
• C. 11
• D. 1

Answer 1

The correct option is A 9

$I_n=(\frac{e^{-x}(sinx{)}^n}{-1}{)}_0^{\infty }+{\int }_0^{\infty }n(sinx{)}^{n-1}{e}^{-x}cosxdx=0+n(\frac{(sinx{)}^{n-1}cosx{e}^{-x}}{-1}{)}_0^{\infty }+n{\int }_0^{\infty }[\left(sinx{)}^{n-1}\right(-sinx)+cox(n-1\left)cox\right(sinx{)}^{n-2}cosx]e^{-x}dx=n{\int }_0^{\infty }[-sinx+(n-1\left)\right(1-si{n}^{2}x\left)\right(sinx{)}^{n-2}]e^{-x}dx=\frac{n(n-1)}{n^2+1}{I}_{n-2}\therefore 101\frac{I_{10}}{I_S}=101\times \frac{10\times 9}{{10}^2+1}=90\therefore \frac{110I_{10}}{I_S}=9$