# Compute the integral of $\left(x^3-1\right)\sin^3\left(\log x\right) \over x^4-x$ from $2$ to $e^\pi$

I tried to compute the following integral: $$\int_2^{e^{\pi}}{\left(x^3-1\right)\sin^3\left(\log x\right) \over x^4-x} \ \mathrm dx=(*)$$

My attempt

$$(*)=\int_2^{e^{\pi}}{\left(x^3-1\right)\sin^3\left(\log x\right) \over x(x^3-1)} \ \mathrm dx=\int_2^{e^{\pi}}{\sin^3\left(\log x\right) \over x} \ \mathrm dx$$ Then I notice that $\frac1{x}=\frac{\mathrm d}{\mathrm dx} \log x$, thus I substituted $\log x=t$: $$=\int_{\log 2}^{\pi}{\sin^3\left(t\right)} \ \mathrm dt$$ Then again another substitution ($\sin t=y$) $$=\int_{\sin(\log 2)}^{\sin\pi}{y^3\over \sqrt{1-y^2}} \ \mathrm dy=-\frac12\int_{0}^{\sin(\log 2)}{y^2\over \sqrt{1-y^2}}2y \ \mathrm dy=$$

Then another substitution ($\sqrt{1-y^2}=u$): $$=\int_{1}^{\sqrt{1-\sin^2(\log 2)}}{1-u^2} \ \mathrm du=-\int^1_{\sqrt{1-\sin^2(\log 2)}}{1-u^2} \ \mathrm du=\left[-u+\frac{u^3}{3}\right]^1_{\sqrt{1-\sin^2(\log 2)}}\simeq-1.284$$

The result given by Wolfram Alpha is $1.284$. Where did I make a mistake?

• Why not write $\sin^3(x)=\sin^2(x)\sin(x)=(1-\cos^2(x))\sin(x)$ and enforce the substitution $t=\cos(x)$? – Mark Viola Mar 24 '18 at 17:51
• It's a faster way to achieve the result, thank you – Lorenzo B. Mar 24 '18 at 18:53

When you're at $\sin^3t$, change it into $\sin t-\sin t\cos^2t$, so the integral is elementary: $$\int_{\log2}^\pi(\sin t-\sin t\cos^2t)\,dt =\Bigl[-\cos t+\frac{\cos^3t}{3}\Bigr]_{\log2}^{\pi}= \frac{2}{3}+\cos\log2-\frac{1}{3}\cos^3\log2\approx1.284$$
You forgot a minus sign, because with $\sqrt{1-y^2}=u$, you have $-2y\,dy=2u\,du$ and you used instead $2y\,dy=2u\,du$.
• Thank you, I noticed the sign mistake I made: I knew that $-2ydy=du$ but instead of putting a minus sign outside and inside the integral sign, I put it only outside – Lorenzo B. Mar 24 '18 at 18:02
An easy alternative can be using an identity $\sin (3x)=3\sin(x)-4\sin^3 (x)$ and plugging $\sin^3 (x)=\frac {3\sin (x)-\sin (3x)}{4}$