$2int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{2m(cos(mpi)+1)}{pi (m^2-1)}$ when $m ge 2 $












3












$begingroup$


$$2int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{2m(cos(mpi)+1)}{pi (m^2-1)},; mge2$$



I can't seem to be able to prove this.



I have tried using the identity $sin(a)cos(b) = 1/2 (sin(a-b)+sin(a+b))$ to get $$int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{1}{2} int_{0}^{1} sin((m-1)pi x) + sin((m+1)pi x) dx$$ and then solving but am unable to reproduce the solution.






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 7 '18 at 14:16










  • $begingroup$
    Hint: linearize $cos asin b$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 7 '18 at 14:17












  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Dec 7 '18 at 14:18










  • $begingroup$
    ive tried using the identity sin(a)cos(b)=1/2(sin(a−b)+sin(a+b)) but have had no success
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:23










  • $begingroup$
    @pablo_mathscobar that's exactly the right idea. So where are you getting stuck?
    $endgroup$
    – Omnomnomnom
    Dec 7 '18 at 14:27
















3












$begingroup$


$$2int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{2m(cos(mpi)+1)}{pi (m^2-1)},; mge2$$



I can't seem to be able to prove this.



I have tried using the identity $sin(a)cos(b) = 1/2 (sin(a-b)+sin(a+b))$ to get $$int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{1}{2} int_{0}^{1} sin((m-1)pi x) + sin((m+1)pi x) dx$$ and then solving but am unable to reproduce the solution.






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 7 '18 at 14:16










  • $begingroup$
    Hint: linearize $cos asin b$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 7 '18 at 14:17












  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Dec 7 '18 at 14:18










  • $begingroup$
    ive tried using the identity sin(a)cos(b)=1/2(sin(a−b)+sin(a+b)) but have had no success
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:23










  • $begingroup$
    @pablo_mathscobar that's exactly the right idea. So where are you getting stuck?
    $endgroup$
    – Omnomnomnom
    Dec 7 '18 at 14:27














3












3








3


1



$begingroup$


$$2int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{2m(cos(mpi)+1)}{pi (m^2-1)},; mge2$$



I can't seem to be able to prove this.



I have tried using the identity $sin(a)cos(b) = 1/2 (sin(a-b)+sin(a+b))$ to get $$int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{1}{2} int_{0}^{1} sin((m-1)pi x) + sin((m+1)pi x) dx$$ and then solving but am unable to reproduce the solution.






calculus integration







share|cite|improve this question











$endgroup$




$$2int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{2m(cos(mpi)+1)}{pi (m^2-1)},; mge2$$



I can't seem to be able to prove this.



I have tried using the identity $sin(a)cos(b) = 1/2 (sin(a-b)+sin(a+b))$ to get $$int_{0}^{1} cos(pi x)sin(mpi x)dx = frac{1}{2} int_{0}^{1} sin((m-1)pi x) + sin((m+1)pi x) dx$$ and then solving but am unable to reproduce the solution.






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 15:54









amWhy

1




1










asked Dec 7 '18 at 14:16









pablo_mathscobarpablo_mathscobar

996




996












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 7 '18 at 14:16










  • $begingroup$
    Hint: linearize $cos asin b$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 7 '18 at 14:17












  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Dec 7 '18 at 14:18










  • $begingroup$
    ive tried using the identity sin(a)cos(b)=1/2(sin(a−b)+sin(a+b)) but have had no success
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:23










  • $begingroup$
    @pablo_mathscobar that's exactly the right idea. So where are you getting stuck?
    $endgroup$
    – Omnomnomnom
    Dec 7 '18 at 14:27


















  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 7 '18 at 14:16










  • $begingroup$
    Hint: linearize $cos asin b$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 7 '18 at 14:17












  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Dec 7 '18 at 14:18










  • $begingroup$
    ive tried using the identity sin(a)cos(b)=1/2(sin(a−b)+sin(a+b)) but have had no success
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:23










  • $begingroup$
    @pablo_mathscobar that's exactly the right idea. So where are you getting stuck?
    $endgroup$
    – Omnomnomnom
    Dec 7 '18 at 14:27
















$begingroup$
What have you tried?
$endgroup$
– MisterRiemann
Dec 7 '18 at 14:16




$begingroup$
What have you tried?
$endgroup$
– MisterRiemann
Dec 7 '18 at 14:16












$begingroup$
Hint: linearize $cos asin b$.
$endgroup$
– Jean-Claude Arbaut
Dec 7 '18 at 14:17






$begingroup$
Hint: linearize $cos asin b$.
$endgroup$
– Jean-Claude Arbaut
Dec 7 '18 at 14:17














$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 7 '18 at 14:18




$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 7 '18 at 14:18












$begingroup$
ive tried using the identity sin(a)cos(b)=1/2(sin(a−b)+sin(a+b)) but have had no success
$endgroup$
– pablo_mathscobar
Dec 7 '18 at 14:23




$begingroup$
ive tried using the identity sin(a)cos(b)=1/2(sin(a−b)+sin(a+b)) but have had no success
$endgroup$
– pablo_mathscobar
Dec 7 '18 at 14:23












$begingroup$
@pablo_mathscobar that's exactly the right idea. So where are you getting stuck?
$endgroup$
– Omnomnomnom
Dec 7 '18 at 14:27




$begingroup$
@pablo_mathscobar that's exactly the right idea. So where are you getting stuck?
$endgroup$
– Omnomnomnom
Dec 7 '18 at 14:27










1 Answer
1






active

oldest

votes


















6












$begingroup$

Your method gives $int_0^1 (sin (m + 1)pi x+sin (m -1)pi x)dx=frac{-1}{pi}bigg(frac{cos (m+1)pi -1}{m+1}+frac{cos (m-1)pi -1}{m-1}bigg)$. As the cosines are both $-cos mpi$, this simplifies to $frac{2m (cos mpi +1)}{pi (m^2-1)}$. As a sanity check, let's take $m=2$, so my claim is that the answer is $frac{8}{3pi}$. Indeed $$2int_0^1cos pi xsin 2pi x dx=2int_0^1 2sin pi xcos^2 pi x dx,$$which by $u=cospi x$ gives $$int_{-1}^1frac{4}{pi}u^2 du=frac{8}{3pi}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    apologies i forgot to write the 2 when typing the equation out my confusion comes from where you say that the cosines are both $-cos(mpi)$? how does that simplify
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:32










  • $begingroup$
    @pablo_mathscobar From the identity $cos (y+pi)=-cos y$.
    $endgroup$
    – J.G.
    Dec 7 '18 at 14:33










  • $begingroup$
    thank you!!!!!!!
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:34











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Your method gives $int_0^1 (sin (m + 1)pi x+sin (m -1)pi x)dx=frac{-1}{pi}bigg(frac{cos (m+1)pi -1}{m+1}+frac{cos (m-1)pi -1}{m-1}bigg)$. As the cosines are both $-cos mpi$, this simplifies to $frac{2m (cos mpi +1)}{pi (m^2-1)}$. As a sanity check, let's take $m=2$, so my claim is that the answer is $frac{8}{3pi}$. Indeed $$2int_0^1cos pi xsin 2pi x dx=2int_0^1 2sin pi xcos^2 pi x dx,$$which by $u=cospi x$ gives $$int_{-1}^1frac{4}{pi}u^2 du=frac{8}{3pi}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    apologies i forgot to write the 2 when typing the equation out my confusion comes from where you say that the cosines are both $-cos(mpi)$? how does that simplify
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:32










  • $begingroup$
    @pablo_mathscobar From the identity $cos (y+pi)=-cos y$.
    $endgroup$
    – J.G.
    Dec 7 '18 at 14:33










  • $begingroup$
    thank you!!!!!!!
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:34
















6












$begingroup$

Your method gives $int_0^1 (sin (m + 1)pi x+sin (m -1)pi x)dx=frac{-1}{pi}bigg(frac{cos (m+1)pi -1}{m+1}+frac{cos (m-1)pi -1}{m-1}bigg)$. As the cosines are both $-cos mpi$, this simplifies to $frac{2m (cos mpi +1)}{pi (m^2-1)}$. As a sanity check, let's take $m=2$, so my claim is that the answer is $frac{8}{3pi}$. Indeed $$2int_0^1cos pi xsin 2pi x dx=2int_0^1 2sin pi xcos^2 pi x dx,$$which by $u=cospi x$ gives $$int_{-1}^1frac{4}{pi}u^2 du=frac{8}{3pi}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    apologies i forgot to write the 2 when typing the equation out my confusion comes from where you say that the cosines are both $-cos(mpi)$? how does that simplify
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:32










  • $begingroup$
    @pablo_mathscobar From the identity $cos (y+pi)=-cos y$.
    $endgroup$
    – J.G.
    Dec 7 '18 at 14:33










  • $begingroup$
    thank you!!!!!!!
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:34














6












6








6





$begingroup$

Your method gives $int_0^1 (sin (m + 1)pi x+sin (m -1)pi x)dx=frac{-1}{pi}bigg(frac{cos (m+1)pi -1}{m+1}+frac{cos (m-1)pi -1}{m-1}bigg)$. As the cosines are both $-cos mpi$, this simplifies to $frac{2m (cos mpi +1)}{pi (m^2-1)}$. As a sanity check, let's take $m=2$, so my claim is that the answer is $frac{8}{3pi}$. Indeed $$2int_0^1cos pi xsin 2pi x dx=2int_0^1 2sin pi xcos^2 pi x dx,$$which by $u=cospi x$ gives $$int_{-1}^1frac{4}{pi}u^2 du=frac{8}{3pi}.$$






share|cite|improve this answer











$endgroup$



Your method gives $int_0^1 (sin (m + 1)pi x+sin (m -1)pi x)dx=frac{-1}{pi}bigg(frac{cos (m+1)pi -1}{m+1}+frac{cos (m-1)pi -1}{m-1}bigg)$. As the cosines are both $-cos mpi$, this simplifies to $frac{2m (cos mpi +1)}{pi (m^2-1)}$. As a sanity check, let's take $m=2$, so my claim is that the answer is $frac{8}{3pi}$. Indeed $$2int_0^1cos pi xsin 2pi x dx=2int_0^1 2sin pi xcos^2 pi x dx,$$which by $u=cospi x$ gives $$int_{-1}^1frac{4}{pi}u^2 du=frac{8}{3pi}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '18 at 14:35

























answered Dec 7 '18 at 14:28









J.G.J.G.

25.2k22539




25.2k22539












  • $begingroup$
    apologies i forgot to write the 2 when typing the equation out my confusion comes from where you say that the cosines are both $-cos(mpi)$? how does that simplify
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:32










  • $begingroup$
    @pablo_mathscobar From the identity $cos (y+pi)=-cos y$.
    $endgroup$
    – J.G.
    Dec 7 '18 at 14:33










  • $begingroup$
    thank you!!!!!!!
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:34


















  • $begingroup$
    apologies i forgot to write the 2 when typing the equation out my confusion comes from where you say that the cosines are both $-cos(mpi)$? how does that simplify
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:32










  • $begingroup$
    @pablo_mathscobar From the identity $cos (y+pi)=-cos y$.
    $endgroup$
    – J.G.
    Dec 7 '18 at 14:33










  • $begingroup$
    thank you!!!!!!!
    $endgroup$
    – pablo_mathscobar
    Dec 7 '18 at 14:34
















$begingroup$
apologies i forgot to write the 2 when typing the equation out my confusion comes from where you say that the cosines are both $-cos(mpi)$? how does that simplify
$endgroup$
– pablo_mathscobar
Dec 7 '18 at 14:32




$begingroup$
apologies i forgot to write the 2 when typing the equation out my confusion comes from where you say that the cosines are both $-cos(mpi)$? how does that simplify
$endgroup$
– pablo_mathscobar
Dec 7 '18 at 14:32












$begingroup$
@pablo_mathscobar From the identity $cos (y+pi)=-cos y$.
$endgroup$
– J.G.
Dec 7 '18 at 14:33




$begingroup$
@pablo_mathscobar From the identity $cos (y+pi)=-cos y$.
$endgroup$
– J.G.
Dec 7 '18 at 14:33












$begingroup$
thank you!!!!!!!
$endgroup$
– pablo_mathscobar
Dec 7 '18 at 14:34




$begingroup$
thank you!!!!!!!
$endgroup$
– pablo_mathscobar
Dec 7 '18 at 14:34


















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