Given \( u=\cos x \) and \( \int(\cos x)^{m+1} \sin x d x=-\frac{(\cos x)^{6}}{6}+c \). What is the value of \( m \) ?

To find the value of \( m \), we can use the given integral: \[ \int(\cos x)^{m+1} \sin x \, dx = -\frac{(\cos x)^{6}}{6} + c. \] First, we can perform a substitution to simplify the integral. Let \( u = \cos x \), which gives us \( du = -\sin x \, dx \) or \( -du = \sin x \, dx \). Thus, the integral becomes: \[ \int u^{m+1} (-du) = - \int u^{m+1} du = -\frac{u^{m+2}}{m+2} + C. \] Next, reverting back to the variable \( u \), we substitute \( u = \cos x \) to get: \[ -\frac{(\cos x)^{m+2}}{m+2} + C. \] Now, we set this equal to the right side of the original equation: \[ -\frac{(\cos x)^{m+2}}{m+2} = -\frac{(\cos x)^{6}}{6}. \] Removing the negative sign from both sides yields: \[ \frac{(\cos x)^{m+2}}{m+2} = \frac{(\cos x)^{6}}{6}. \] At this point, we can equate the coefficients of \((\cos x)\) since they must be equal for all \( x \). This gives us: \[ m + 2 = 6. \] Solving for \( m \) gives: \[ m = 6 - 2 = 4. \] Thus, the value of \( m \) is: \[ \boxed{4}. \]