# Answer mismatch in definite integral question:$\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$

I faced a problem while evaluating the following integral.

$$\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$$

MY ATTEMPT: Let $$I=\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$$

By King's Rule $$I=-\int_{0}^{2\pi}{\frac{(2\pi-x)^2\sin(x)}{8+\sin^2(x)}}dx$$

On adding the above two integrals,$$2I=\int_{0}^{2\pi}{\frac{(-4{\pi}^2+4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$ or, $$2I=\int_{0}^{2\pi}{\frac{(-4{\pi}^2)\sin(x)}{8+\sin^2(x)}}dx+\int_{0}^{2\pi}{\frac{(4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$ The first integral follows the property $f(x)=-f(2\pi-x)$ so it vanishes. Thus, $$2I=\int_{0}^{2\pi}{\frac{(4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$ or, $$I=2 \pi\int_{0}^{2\pi}{\frac{ x\sin(x)}{8+\sin^2(x)}}dx$$ Applying king's rule and adding, $$2I=2 \pi\int_{0}^{2\pi}{\frac{ 2\pi\sin(x)}{8+\sin^2(x)}}dx$$ $$I=2{\pi}^2\int_{0}^{2\pi}{\frac{ \sin(x)}{8+\sin^2(x)}}dx$$ $$I=2{\pi}^2\int_{0}^{2\pi}{\frac{ \sin(x)}{9-\cos^2(x)}}dx$$

$$I=[2\pi^2(\frac{1}{6} \log(3-\cos(x))-\frac{1}{6} \log(\cos(x)+3))]_{0}^{2\pi}+C$$ Thus,$I=0$ But the answer is not 0.Where did I go wrong?

P.S King's Rule is $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$

• When you apply the rule second time shouldn't it be $(2\pi-x)$ – Archis Welankar Aug 5 '16 at 5:08
• I never knew that was called King's Rule. I did not even know that the integral property had a name – imranfat Aug 5 '16 at 5:21
• @ArchisWelankar i get it now!I should rather use integration by parts! – user220382 Aug 5 '16 at 6:23
• @imranfat See this quora.com/What-is-King-property-of-Integration – user220382 Aug 5 '16 at 6:25
• @SanchayanDutta I made up my own name for this property: Reflexive Integral Property, abbreviated as RIP . I couldn't get a better name :) – imranfat Aug 5 '16 at 14:38

In your second application of King's rule, you forgot the negative sign! $$I=2\pi\int_0^{2\pi}\frac{x \sin(x)}{8+\sin^2(x)}dx$$ becomes $$I=-2\pi\int_0^{2\pi}\frac{(2\pi-x)\sin(x)}{8+\sin^2(x)}dx$$, so adding these two gives you nothing.

Symmetry is a bit more evident if we set $x=z+\pi$. Since $\sin(z+\pi)=-\sin(z)$, $$I = -\int_{-\pi}^{\pi}\frac{(z^2+2\pi z+\pi^2)\sin(z)}{8+\sin(z)^2}\,dz=-2\pi\int_{-\pi}^{\pi}\frac{z\sin(z)}{8+\sin(z)^2}=-4\pi\int_{0}^{\pi}\frac{z\sin z}{8+\sin(z)^2}\,dz$$ and since $\sin(\pi-z)=\sin(z)$, $$I = -4\pi^2\int_{0}^{\frac{\pi}{2}}\frac{\sin z}{8+\sin(z)^2}\,dz=4\pi^2\int_{0}^{\frac{\pi}{2}}\frac{\cos' z}{9-\cos(z)^2}\,dz =-4\pi^2\int_{0}^{1}\frac{du}{9-u^2}$$ and by partial fraction decomposition: $$I = -\frac{2\pi^2}{3}\int_{0}^{1}\left(\frac{1}{3-u}+\frac{1}{3+u}\right)\,du=-\frac{2\pi^2}{3}\left(\log\frac{3}{2}+\log\frac{4}{3}\right)=\color{red}{-\frac{2\pi^2}{3}\log 2}.$$

The second time apply King's rule and add you do so incorrectly due to a plus/minus error. The $x$ term should not have cancelled out.

$$I=2 \pi\int_{0}^{2\pi}{\frac{ x\sin(x)}{8+\sin^2(x)}}dx$$

Applying the rule gives:

$$I=2 \pi\int_{0}^{2\pi}{\frac{ (2\pi-x)\sin(2\pi-x)}{8+\sin^2(2\pi-x)}}dx$$

$$I=-2 \pi\int_{0}^{2\pi}{\frac{ (2\pi-x)\sin(x)}{8+\sin^2(x)}}dx$$

$$=2 \pi\int_{0}^{2\pi}{\frac{ (x-2\pi)\sin(x)}{8+\sin^2(x)}}dx$$

$$2I=2 \pi\int_{0}^{2\pi}{\frac{ (2x-2\pi)\sin(x)}{8+\sin^2(x)}}dx$$