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 $$ m $$ ?

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To determine the value of $$ m $$, let's analyze the given integral:

$$
\int (\cos x)^{m+1} \sin x \, dx = -\frac{(\cos x)^6}{6} + C
$$

Given the substitution $$ u = \cos x $$, we find:
$$
du = -\sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -du
$$

Substituting into the integral:
$$
\int (\cos x)^{m+1} \sin x \, dx = \int u^{m+1} (-du) = -\int u^{m+1} \, du
$$

Integrating:
$$
-\int u^{m+1} \, du = -\frac{u^{m+2}}{m+2} + C = -\frac{(\cos x)^{m+2}}{m+2} + C
$$

Comparing this with the given expression:
$$
-\frac{(\cos x)^6}{6} + C
$$

We see that the exponents and coefficients must match:
$$
m + 2 = 6 \quad \Rightarrow \quad m = 4
$$
$$
m + 2 = 6 \Rightarrow \text{Coefficient: } \frac{1}{m+2} = \frac{1}{6}, \text{ which is consistent.}
$$

**Answer:** $$ m = 4 $$
$$ m = 4 $$

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