Question

Sorry, one small doubt:

\[\frac{2}{T}\int\limits_{0}^{\frac{T}{2}} \sin( \omega t) \sin (k \omega t) \cdot dt = \frac{1}{T} \int\limits_{0}^{\frac{T}{2}}\cos( \omega t(n-1)) - \cos( \omega t(n+1)) \cdot dt\]

How this integral leads to :

= \(\begin{cases} \frac{1}{2} , && n = 1 && \\0, && n \ge 2, && n \in \mathbb{N} \end{cases}\)

Answer 1

\[\omega = \frac{2 \pi}{T}\]

Answer 2

Let us put \(n = 1\) directly..

Answer 3

First term becomes \(1\)

Answer 4

Second term becomes:

\[\cos(2 \omega t) = \cos(\frac{4 \pi t}{T})\]

Answer 5

Where I am misinterpreting it?? :(

Answer 6

integral cos(w(n-1)*t) = sin(w(n-1)*t)/(w(n-1)) limit

Answer 7

If you will integrate it, then you will get all Integral to be \(0\)..

Answer 8

yeah

Answer 9

for n>=2 it is 0

Answer 10

u have prob at n=1??

Answer 11

Not for \(n=1\) ??

Answer 12

integrate it

Answer 13

Wait.. :)

Answer 14

sin(4pi*t/T)/(4pi/T)

Answer 15

\[\frac{1}{T} \int\limits\limits_{0}^{\frac{T}{2}}\cos( \omega t(n-1)) - \cos( \omega t(n+1)) \cdot dt = \frac{1}{T} [\frac{\sin(\omega t(n-1))}{ \omega (n-1)} - \frac{\sin(\omega t(n+1))}{ \omega (n+1)}]_{0}^{T/2}\]

Answer 16

\[ \frac{1}{T} [\frac{\sin(\omega t(n-1))}{ \omega (n-1)} - \frac{\sin(\omega t(n+1))}{ \omega (n+1)}]_{0}^{T/2}\]

Answer 17

Is it?

Answer 18

eh question why did you go from k to n? in your first integral

Answer 19

yeah ....go further

Answer 20

put limits

Answer 21

Sorry, that is not \(k\) but \(n\).. you can take \(k=n\)..

Answer 22

for k=1
put in start k=1

Answer 23

sinwt*sinwt

Answer 24

sin^2wt
=1-cos2wt

Answer 25

integral
t-sin2wt/2w

Answer 26

eh okay!

Answer 27

\[\frac{ T }{ 2 }-\frac{ \sin \frac{ 2*\pi }{ T} \frac{ T }{ 2 }}{ \frac{ 2*\pi }{ T } }\]

Answer 28

sin term will be zero

Answer 29

so we left with T/2 and 2?t is outside integral

Answer 30

\[\frac{ 2 }{ T }*\frac{ T }{ 2 }=1\]

Answer 31

oh sorry paaji

Answer 32

\[\frac{1}{T} [\frac{\sin(\omega t(n-1))}{ \omega (n-1)} - \frac{\sin(\omega t(n+1))}{ \omega (n+1)}]_{0}^{T/2} = \frac{1}{ T \omega}[\frac{\sin( \pi (n-1))}{n-1} - \frac{\sin( \pi (n+1))}{n+1}]\]

Answer 33

cos2x=1-2sin^2x
2sin^2x=1-cos2x

Answer 34

so 2 we took inside in the start ....i missed that

Answer 35

so ans = 1/2

Answer 36

@gorv

Answer 37

I want to get this result from Integration result not from looking at the integral and using your brain.. :P

Answer 38

well u yourself said let n=1 put it first

Answer 39

take the intergral further

Answer 40

now we need to place the value for n agree??

Answer 41

so at n=1 n-1 =0

Answer 42

n-1 is in denominator of one term

Answer 43

so u can not put dierctly n=1 u need to apply limit

Answer 44

term with denominator ( n+1) at n=1 is =0

Answer 45

\[\lim_{n \rightarrow 1}\frac{ 1 }{ T*w } \frac{ \sin (\pi*(n-1) }{ n-1 }\]

Answer 46

w=2pi/T
wT=2*pi

Answer 47

\[\frac{ 1 }{ 2 } \lim_{n \rightarrow 1}\frac{ \sin(\pi*(n-1) }{ \pi*(n-1) }\]

Answer 48

=1/2 hence we also proved it by integral

Answer 49

\[\lim_{\theta \rightarrow 0}\frac{ \sin \theta }{ \theta }=1\]

Answer 50

@waterineyes

Answer 51

Oh, my mind was not thinking of this.. :)

Thanks paaji.. :)

Answer 52

welcome paaji :)