If $\int \cos ^7 x d x=A \sin ^7 x+B \sin ^5 x+C \sin ^3 x+\sin x+k$, then

$A=\frac{1}{7}, B=\frac{3}{5}, C=-1$ $A=-\frac{1}{7}, B=\frac{3}{5}, C=-1$ $A=\frac{-1}{7}, B=\frac{1}{5}, C=-1$ $A=\frac{1}{7}, B=\frac{3}{5}, C=1$

$A=-\frac{1}{7}, B=\frac{3}{5}, C=-1$

We have, $I=\int \cos ^7 x d x=\int\left(1-\sin ^2 x\right)^3 d(\sin x)$ $\Rightarrow I =\int\left(1-3 \sin ^2 x+3 \sin ^4 x-\sin ^6 x\right) d(\sin x)$ $\Rightarrow I =\sin x-\sin ^3 x+\frac{3}{5} \sin ^5 x-\frac{1}{7} \sin ^7 x+k$ ∴  $A=-\frac{1}{7}, B=\frac{3}{5}$ and $C=-1$