Given $$I = \int\limits_0^\pi \frac{x\sin x}{1+\cos^2\!x}dx$$
prove without calculating $ \frac{\pi}{2} \le I \le \pi $
calulate $I$
so far I know $I = \int\limits_0^\pi \frac{x\sin x}{1+\cos^2\!x}dx \le \int\limits_0^\pi x\sin x =\pi$
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jgon
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Daniel Katzan
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2Your thoughts? Your efforts? – Vincenzo Oliva Aug 12 '15 at 08:44
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Hint: integrate by parts ($I = -\int x,d(\arctan(\cos x))$) and look at graph of $\arctan(\cos x)$ – Michael Galuza Aug 12 '15 at 09:23
2 Answers
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We have $$I = \int\limits_0^\pi \frac{x\sin x}{1+\cos^2\!x}dx\tag 1$$
Using property of definite integral $\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$, we get $$I = \int\limits_0^\pi \frac{(\pi-x)\sin (\pi-x)}{1+\cos^2(\pi-x)}dx$$ $$I = \int\limits_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2 x}dx\tag 2$$ Now, adding (1) & (2), we get $$I+I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^2 x}dx$$ $$I=\frac{\pi}{2}\int_{0}^{\pi}\frac{\sin x}{1+\cos^2 x}dx$$
$$I=-\frac{\pi}{2}\left[\tan^{-1}(\cos x)\right]_{0}^{\pi}$$ $$I=-\frac{\pi}{2}\left[\tan^{-1}(\cos \pi)-\tan^{-1}(\cos 0)\right]$$ $$I=-\frac{\pi}{2}\left[\tan^{-1}(-1)-\tan^{-1}(1)\right]$$ $$I=-\frac{\pi}{2}\left[\frac{-\pi}{4}-\frac{\pi}{4}\right]=\frac{\pi^2}{4}$$
Harish Chandra Rajpoot
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@harischandrarajppot. How would you show without calculating that it is more than half $\pi$ ? That is also part o of question 1. – Shailesh Aug 12 '15 at 09:03
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@VincenzoOliva No. The OP was to show that without evaluating the integral. – Shailesh Aug 12 '15 at 10:18
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@Shailesh: Oh, I see, misunderstood you. Indeed I had thought of commenting like you, but I desisted when the OP accepted this answer. – Vincenzo Oliva Aug 12 '15 at 10:20
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@VincenzoOliva. To me the problem is still not fully solved. It may be very straightforward to show the remaining part, but the answer needs to address that issue. – Shailesh Aug 12 '15 at 10:32
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1A more generic formula : $$\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$ – lab bhattacharjee Aug 12 '15 at 10:32
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1
$I=\int_0 ^\pi$$\frac{xsinx}{1+cos^2 x}$$dx$....(i)
$\Rightarrow I= \int_0 ^\pi \frac{(\pi-x)sinx}{1+cos^2x} dx$...(ii)
Summing (i) and (ii) we have,
$2I=\pi\int_0 ^\pi \frac{sinx}{1+cos^2x}dx$
$\Rightarrow I=\frac \pi2 \int_0 ^\pi \frac{sinx}{1+cos^2x}dx$
Now assume that $cosx=z$ $\Rightarrow -sinx=dz$
Now replace the variables and the according limits in your integral
i.e. $I= \frac \pi2 \int_0 ^\pi \frac{-z}{1+z^2}dz$
Then it is nothing difficult to find the result.