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

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 $C=-1$