Question

lf $I_n=\int co{s}^{n}xdx$ then $I_7-\frac{co{s}^{6}xsinx}{7}=$

• A. $\frac{6}{7}{I}_5$
• B. $-\frac{6}{7}{I}_5$
• C. $\frac{5}{8}{I}_5$
• D. $-\frac{5}{8}{I}_5$

Answer 1

The correct option is A $\frac{6}{7}{I}_5$

$I_n-\int co{s}^{n}xdx\Rightarrow In=\int co{s}^{n+2}xse{c}^{2}xdx$

$=co{s}^{n-2}x\int se{c}^{2}dx+\int co{s}^{n+1}x(n+2)(+sinx)tanxdx$

$I_n=co{s}^{n+2}xtanx+(n+2)\int co{s}^n\left(si{n}^{2}x\right)dx$

$I_n=co{s}^{n+2}tanx+(n+2)\{[\int co{s}^{n}dx]-\int co{s}^{n+2}dx\}$

$I_n=co{s}^{n+2}tanx+(n+2)[I_n-I_{n+2}]$

$(n+2)(In+2)-co{s}^{n+2}tanx=(n+2)I_n-I_n$

= let n = 5

$7I_7-co{s}^{6}sinx=+6In$

$I_7-\frac{co{s}^{6}xsinx}{7}={}^6{/}_7{I}_n$