# You need to prove that question int_0^1 sin (pimx)sin(pinx)dx={(0m !ne n),(1/2m=n):}

Khaleesi Herbert 2021-01-19 Answered

You need to prove that question
${\int }_{0}^{1}\mathrm{sin}\left(\pi mx\right)\mathrm{sin}\left(\pi nx\right)dx=\left\{\left(\begin{array}{c}0m\ne n\\ 1\text{/}2m=n\end{array}\right)$

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Szeteib
We have to show
${\int }_{0}^{1}\mathrm{sin}\left(\pi mx\right)\mathrm{sin}\left(\pi nx\right)dx=\left\{\left(\begin{array}{c}0m\ne n\\ 1\text{/}2m=n\end{array}\right)$
We will use the following trigonometric identities:
$\mathrm{sin}A\mathrm{sin}B=\frac{1}{2}\left[\mathrm{cos}\left(A-B\right)-\mathrm{cos}\left(A+B\right)\right]$
Thus we have
$\mathrm{sin}\left(\pi mx\right)\mathrm{sin}\left(\pi nx\right)=\frac{1}{2}\left[\mathrm{cos}\left(m-n\right)\pi x-\mathrm{cos}\left(m+n\right)\pi x\right]$
First consider the case $m\ne n$. Then we have
${\int }_{0}^{1}\mathrm{sin}\left(\pi mx\right)\mathrm{sin}\left(\pi nx\right)dx={\int }_{0}^{1}\frac{1}{2}\left[\mathrm{cos}\left(m-n\right)\pi x-\mathrm{cos}\left(m+n\right)\pi x\right]dx$
$=\left(\frac{\mathrm{sin}\left[\left(m-n\right)\pi x\right]}{2\left(m-n\right)\pi }-\frac{\mathrm{sin}\left[\left(m+n\right)\pi x\right]}{2\left(m+n\right)\pi }\right){|}_{0}^{1}$
$=0,$
because $\mathrm{sin}\left(k\pi \right)=0$, for all k in ZZ. Considering $m=n$ than we have
${\int }_{0}^{1}{\mathrm{sin}}^{2}\left(\pi mx\right)dx={\int }_{0}^{1}\frac{1}{2}\left[1-\mathrm{cos}2m\pi x\right]dx$
$=\left(\frac{x}{2}-\frac{\mathrm{sin}\left[2m\pi x\right]}{4m\pi }\right){|}_{0}^{1}$
$=\frac{1}{2},$
And we get the finally answer is
${\int }_{0}^{1}\mathrm{sin}\left(\pi mx\right)\mathrm{sin}\left(\pi nx\right)dx=\left\{\left(\begin{array}{c}0m\ne n\\ 1\text{/}2m=n\end{array}\right)$