Question

If $\int cose{c}^{n}xdx=\frac{cose{c}^{n-2}x\cot x}{n-1}+f\left(n\right)\int cose{c}^{n-2}xdx$, then$f(n+1)$ is equal to

• A. $\frac{n-2}{n-1}$
• B. $\frac{n-1}{n}$
• C. $\frac{n-3}{n-2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{n-1}{n}$

Given, $\int cose{c}^{n}xdx=\frac{cose{c}^{n-2}x\cot x}{n-1}+f\left(n\right)\int cose{c}^{n-2}xdx$
$I_n=\int cose{c}^{n}xdx=\int cose{c}^{n-2}xcose{c}^{2}xdx$
$=\frac{cose{c}^{n-2}x\cot x}{-1}-(n-2)\int cose{c}^{n-3}x(-cosecx\cot x)(-\cot x)dx$
$=\frac{cose{c}^{n-2}x\cot x}{-1}-(n-2)\int cose{c}^{n-2}x(cose{c}^{2}x-1)dx$
$=\frac{cose{c}^{n-2}x\cot x}{-1}-(n-2)[\int cose{c}^{n}xdx-\int cose{c}^{n-2}xdx]$
$I_n=\frac{cose{c}^{n-2}x\cot x}{-1}-(n-2)I_n+(n-2)I_{n-2}$
$(n-1)I_n=\frac{cose{c}^{n-2}x\cot x}{-1}+(n-2)I_{n-2}$
$\therefore$ $I_n-\frac{cose{c}^{n-2}x\cot x}{n-1}+\frac{n-2}{n-1}{I}_{n-2}$
$I_n=-\frac{cose{c}^{n-2}x\cot x}{n-1}+f\left(n\right)\int cose{c}^{n-2}xdx$
where $f\left(n\right)=\frac{n-2}{n-1}$

$\therefore f(n+1)=\frac{n-1}{n}$
Hence, option 'B' is correct.