To evaluate the integral \(\int \cos^7(\theta) \sin(\theta) , d\theta\) using the substitution \(u = \cos(\theta)\), we can follow these steps:
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Substitution: We have \[ u = \cos(\theta) \implies du = -\sin(\theta) , d\theta \quad \text{or} \quad d\theta = -\frac{du}{\sin(\theta)}. \] Since \(\sin(\theta) = \sqrt{1 - u^2}\) (using the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\)), we can express \(\sin(\theta)\) in terms of \(u\).
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Rewrite the integral: The sine term can be used to rewrite \(d\theta\): \[ d\theta = -\frac{du}{\sqrt{1 - u^2}}. \]
Therefore, the integral becomes: \[ \int \cos^7(\theta) \sin(\theta) , d\theta = \int u^7 \sin(\theta) , d\theta = \int u^7 \sin(\theta) \left(-\frac{du}{\sin(\theta)}\right). \] As \(\sin(\theta)\) cancels, we have: \[ = - \int u^7 , du. \]
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Integrate: Now, we can integrate: \[ -\int u^7 , du = -\left(\frac{u^8}{8}\right) + C = -\frac{u^8}{8} + C. \]
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Back-substituting: Recall that \(u = \cos(\theta)\), so substituting back gives us: \[ -\frac{(\cos(\theta))^8}{8} + C = -\frac{\cos^8(\theta)}{8} + C. \]
Therefore, the final answer is: \[ \int \cos^7(\theta) \sin(\theta) , d\theta = -\frac{\cos^8(\theta)}{8} + C. \]