Fourier series Basics

The value of sin $n\pi$  is

• 1

• 0

• -1

• n

The value of cos $n\pi=$  ------ and  Cos 0=-------

• 0 and 0

•  $\left(-1\right)^n$   and 0

•  $\left(-1\right)^n$  and 1

• 1 and 1

The value of cos $2\pi$  and cos $\pi$  

• 0 and 0

• 1 and 1

• 1 and -1

• -1 and -1

The value of  $\int_{ }^{ }\cos5x\ dx$  =

• 5 sin5x

• -5sin5x

•  $\frac{\sin5x}{5}$  

•  $-\frac{\sin5x}{5}$  

The value of $\int_0^{\pi}\ \sin3x\ dx$  

•  $\frac{2}{3}$  

•  $-\frac{2}{3}$  

• 0

•  $\frac{1}{3}$  

 $\frac{\text{d}}{\text{d}x}\left(x^2\right)\ =$  

• x

• 2

• 2x

• 0

 $\frac{\text{d}}{\text{d}x}\left(k\right)=$  -------where k is constant

• 1

• k

• 0

• x

 $\int_0^{2\pi}\ \ x\ \sin x\ dx$  =

•  $\left|x\left(-\cos x\right)-\left(-\sin x\right)\right|$  

•  $\left|x\left(\cos x\right)-\left(\sin x\right)\right|$  

•  $\left|x\left(-\cos x\right)-\left(-\sin x\right)\right|_0^{2\pi}\ =\ 0$  

•  $\left|x\left(-\cos x\right)-\left(-\sin x\right)\right|_0^{2\pi}\ =\ -2\pi$  

 $\int_0^{2\pi}\ x\ \cos nx\ dx\ =$  

•  $\left|x\left(\sin x\right)\left(-\cos x\right)\right|$  

•  $\left|x\left(\frac{\sin nx}{n}\right)-\left(-\frac{\cos nx}{n^2}\right)\right|$  

•  $\left|x\left(\frac{\sin nx}{n}\right)-\left(-\frac{\cos nx}{n^2}\right)\right|_0^{2\pi}\ \ =0$  

•  $\left|x\left(\frac{\sin nx}{n}\right)-\left(-\frac{\cos nx}{n^2}\right)\right|_0^{2\pi}\ =2$  

 $\int_0^{2\pi}\ x\ \cos\left(\frac{n\pi}{l}\ \right)x\ dx\ =$  

•  $\left|x\left(\sin\left(\frac{n\pi}{l}\ \right)x\right)\left(-\cos\left(\frac{n\pi}{l}\ \right)x\right)\right|$  

•  $\left|x\left(\frac{\sin\left(\frac{n\pi}{l}\ \right)x}{\left(\frac{n\pi}{l}\ \right)}\right)-\left(-\frac{\cos\left(\frac{n\pi}{l}\ \right)x}{\left(\frac{n\pi}{l}\ \right)^2}\right)\right|$  

•  $\left|x\left(\frac{\sin\left(\frac{n\pi}{l}\ \right)x}{\left(\frac{n\pi}{l}\ \right)}\right)-\left(-\frac{\cos\left(\frac{n\pi}{l}\ \right)x}{\left(\frac{n\pi}{l}\ \right)^2}\right)\right|_0^{2\pi}\ \$                           
  
      = 0

•  $\left|x\left(\frac{\sin nx}{n}\right)-\left(-\frac{\cos nx}{n^2}\right)\right|_0^{2\pi}\$  
   =2