Integrating $ -\frac{A}{4}\int_0^T \sin\left(\frac{2\pi t}{T} - \frac{4\pi \tau}{T}\right)d\tau $

Question

Let $T$ and $A$ be constants, how do I solve the following integral? $$ -\frac{A}{4}\int_0^T \sin\left(\frac{2\pi t}{T} - \frac{4\pi \tau}{T}\right)d\tau $$ The solution of the integral has to be 0.

I have tried to substitute so that the term is equal to $-\frac{A}{4}\int_0^T \sin(u) \left(\frac{T\tau}{2\pi t} - \frac{T}{4\pi}\right) du$

But even if this correct, then I would have to substitute again, so this cannot be the right way.

Answer 1

First notice that $\displaystyle\int_0^{2\pi}\sin(u)du=0\quad$ (same if it was cosinus under the integral).

Shifting the variable doesn't change the value:

$\require{cancel}\displaystyle\begin{align}\int_0^{2\pi}\sin(u+\varphi)du &=\int_{\varphi}^{2\pi+\varphi}\sin(u)du\\ &=\int_{\varphi}^{0}\sin(u)du+\int_{0}^{2\pi}\sin(u)du+\int_{2\pi}^{2\pi+\varphi}\sin(u)du\\ &=-\cancel{\int_0^{\varphi}\sin(u)du}+\int_{0}^{2\pi}\sin(u)du+\cancel{\int_{0}^{\varphi}\underbrace{\sin(u+2\pi)}_{\sin(u)}du}\\ &=\int_{0}^{2\pi}\sin(u)du\end{align}$

And this stays the same if you take the integral over an integer number of periods (I let it as an exercise for you):

$$\int_{0}^{2\pi}\sin(nu+\phi)du=0\quad n\in\mathbb Z^*$$

In the present case $\varphi=\dfrac{2\pi t}T$ is a constant.

And for $\tau\in[0,T]$ then $\dfrac{4\pi\tau}{T}\in[0,4\pi]$ covers $n=2$ times the period.

Therefore your integral is zero without having to do any calculation!

Answer 2

$$ \begin{split} \frac{A}{4}\int_0^T \sin\left( \frac{4\pi}{T}\left(\tau -\frac{t}{2}\right)\right)\, \mathrm{d}\tau &\overset{u = \frac{4\pi}{T}\left(\tau -\frac{t}{2}\right)}{=} \frac{AT}{16\pi} \int_{-\frac{2t\pi}{T}}^{\frac{4\pi}{T}\left(T-\frac{t}{2}\right)}\sin(u) \,\mathrm{d}u \\ &= \frac{AT}{16\pi} \cos(u)\Bigg\vert_{u = {\frac{4\pi}{T}\left(T-\frac{t}{2}\right)}}^{u=-\frac{2t\pi}{T}} \end{split} $$

Answer 3

Let $x =\frac{2 \pi t}{T}-\frac{4 \pi \tau}{T} \textrm{ then } d x =-\frac{4 \pi}{T} d \tau $ and $$ \begin{aligned} -\frac{A}{4}\int_0^T \sin\left(\frac{2\pi t}{T} - \frac{4\pi \tau}{T}\right)d\tau &=-\frac{A}{4} \int_{\frac{2 \pi t}{T}}^{\frac{2 \pi t}{T} -4\pi} \sin x \left(-\frac{4 \pi}{T} d \tau\right) \\ &=-\frac{A \pi}{T}\left[\cos x\right]_{\frac{2 \pi t}{T}}^{\frac{2 \pi t}{T}-4 \pi} \\ &=-\frac{A \pi}{T}\left[\cos \left(\frac{2 \pi t}{T}-4 \pi\right)-\cos \frac{2 \pi t}{T}\right] \\ &=-\frac{A \pi}{T}\left(\cos \frac{2 \pi t}{T}-\cos \frac{2 \pi t}{T}\right) \\ &=0 \end{aligned} $$ Wish it helps.