I am not really sure how to approach this. I have tried to use integration by parts but it does not seem to work.
Thank you
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First off, we can get rid of the $i$ factor from both sides, since $i\ne 0$, this is an equivalent transformation. – Daniel P Mar 02 '22 at 17:58
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This seems to not hold, at least not for any $f$. If $f$ is simply the identity, then the left side is $i(\sin(p)-p\cos(p))$, and the right side is $\frac{i}{2}\left(1-\cos(p)\right)$, which are different functions. (For one, the left side results in an even function, the right side results in an odd function.) – Daniel P Mar 02 '22 at 18:03
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I think you mean to have $\pi$ as the upper limit on both integrals, with the factor $\pi$ in the right-hand side. This is a well-known identity which is easily proved by substituting $u=\pi-t$ – David Quinn Mar 02 '22 at 18:48
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@DavidQuinn yes that's what I meant - I am sorry for the typo - I am not very familiar with how to get the pi symbol. So while substituting, do you have to split the integral and then prove? – a22 Mar 03 '22 at 18:52
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@DanieP - I just realised that there was a typo here - i meant for the upper limit to equal pi - I am very new to the platform so I was unsure how to get the pi symbol - sorry for that – a22 Mar 03 '22 at 18:53
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type \pi with to get $\pi$. No just substitute $u=\pi-t$ – David Quinn Mar 03 '22 at 18:54
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@akr22 $\sin(t)$ won't change under this substitution. – Jakobian Mar 03 '22 at 19:02
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@DavidQuinn Thank you, so regarding the substitution do you then rearrange for t and then replaced the t with pi - u to get the integral (pi to 0) [ (pi - u)f(sin(u))] = 1/2 integral (pi to 0) f(sinu)] ? – a22 Mar 03 '22 at 19:04
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@Jakobian - it would just be sin(u) right which is just sin(t)? – a22 Mar 03 '22 at 19:04
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Does this answer your question? Let $A=\pi^2\int_{0}^{1}\frac{\sin(\pi x)}{1 + \sin(\pi x)}dx$ and $B=\int_{0}^{\pi}\frac{x\sin( x)}{1 + \sin( x)}dx$. Find $\frac{A}{B}$. – Robert Lee Mar 07 '22 at 06:30
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@RobertLee Not really because I am not sure how to integrate the tf(sint) bit – a22 Mar 10 '22 at 12:24
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From left side
$$I=\int_{0}^{\pi}xf(\sin x)dx$$
$$=\int_{0}^{\pi}(\pi-x)f(\sin x)dx$$
hence $$\color{red}I= \frac{\pi}{2} \color{blue}\int_{0}^{\pi}f(\sin x)dx$$
Surjeet Singh
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