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I have a problem solving an integration. This is my approach: \begin{align*} u &= \frac{1}{T}\cdot\intop_{0}^{T}(U_{0}+\hat{u}\cdot \sin(\omega t))dt\qquad\text{with }\omega=\frac{2\pi}{T}\\ u &= \frac{1}{T}\cdot\intop_{0}^{T}(U_{0}+\hat{u}\cdot \sin(\frac{2\pi}{T}\cdot t))dt\\ u &= \frac{1}{T}\cdot\left[U_{0}t-\hat{u}\cdot\frac{T}{2\pi}\cdot\cos\frac{2\pi}{T}\cdot t\right]_{0}^{T}\\ u &= \frac{U_{0}T}{T}-\hat{u}\cdot\frac{T}{2\pi\cdot T}\cdot\cos\frac{2\pi\cdot T}{T}\\ u &= U_{0}-\hat{u}\cdot\frac{1}{2\pi}\cdot\cos2\pi\\ u &= U_{0}-\frac{\hat{u}}{2\pi}. \end{align*}

But the solution should be ${u = U_{0}}$. Where is my mistake?

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up vote 6 down vote accepted

It's $\cos 0=1$, not $\cos 0=0$.

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What do you mean? I only have $\cos(2\pi)=1$ – Sven Walter Feb 12 '11 at 14:02
You are missing the term that comes from inserting the lower limit $t=0$ into the antiderivative. That term will cancel the one coming from $t=T$, since $\cos 2\pi$ and $\cos 0$ are both 1. – Hans Lundmark Feb 12 '11 at 14:11
Oh dear! I didn't see that. Thanks! – Sven Walter Feb 12 '11 at 14:19

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