Expert answers to "You need to prove that question ∫01sin⁡(πmx)sin⁡(πnx)dx={(0m≠n1/2m=n)"

Khaleesi Herbert

Answered question

2021-01-19

You need to prove that question
$\int_{0}^{1} \sin (π m x) \sin (π n x) d x = {(\begin{matrix} 0 m \neq n \\ 1 / 2 m = n \end{matrix})$

Szeteib

Skilled2021-01-20Added 102 answers

We have to show

\int_{0}^{1} \sin (π m x) \sin (π n x) d x = {(\begin{matrix} 0 m \neq n \\ 1 / 2 m = n \end{matrix})

We will use the following trigonometric identities:

\sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)]

Thus we have

\sin (π m x) \sin (π n x) = \frac{1}{2} [\cos (m - n) π x - \cos (m + n) π x]

First consider the case

m \neq n

. Then we have

\int_{0}^{1} \sin (π m x) \sin (π n x) d x = \int_{0}^{1} \frac{1}{2} [\cos (m - n) π x - \cos (m + n) π x] d x

= (\frac{\sin [(m - n) π x]}{2 (m - n) π} - \frac{\sin [(m + n) π x]}{2 (m + n) π}) |_{0}^{1}

= 0,

because

\sin (k π) = 0

, for all k in ZZ. Considering

m = n

than we have

\int_{0}^{1} \sin^{2} (π m x) d x = \int_{0}^{1} \frac{1}{2} [1 - \cos 2 m π x] d x

= (\frac{x}{2} - \frac{\sin [2 m π x]}{4 m π}) |_{0}^{1}

= \frac{1}{2},

And we get the finally answer is

\int_{0}^{1} \sin (π m x) \sin (π n x) d x = {(\begin{matrix} 0 m \neq n \\ 1 / 2 m = n \end{matrix})

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