Calculus Problem No.5

The substitution $y = -x$ tells us that $I \; = \; \int_{-\frac16\pi}^{\frac16\pi} \frac{x \cos x}{\sqrt{1 + x^2} + x}\,dx \; = \; \int_{-\frac16\pi}^{\frac16\pi} \frac{-y \cos y}{\sqrt{1+y^2}-y}\,dy$ and hence

$\begin{aligned} I & = \; \frac12\left(\int_{-\frac16\pi}^{\frac16\pi} \frac{x \cos x}{\sqrt{1 + x^2} + x}\,dx - \int_{-\frac16\pi}^{\frac16\pi} \frac{x \cos x}{\sqrt{1+x^2}-x}\,dx \right) \\[2ex] & = \; \frac12\int_{\frac16\pi}^{\frac16\pi} \frac{x\big[\big(\sqrt{1+x^2}-x) - \big(\sqrt{1+x^2} + x\big)\big]\cos x}{\sqrt{1+x^2}^2 - x^2}\,dx \\ & = \; -\int_{-\frac16\pi}^{\frac16\pi} x^2 \cos x\,dx \; = \; -\Big[x^2\sin x + 2x \cos x - 2\sin x\Big]_{-\frac16\pi}^{\frac16\pi} \; =\; \boxed{2 - \tfrac{\pi}{\sqrt{3}} - \tfrac{\pi^2}{36}} \end{aligned}$

$I=\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{x\cos x}{\sqrt{1+x^2}+x} \, dx=\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{x \cos x (\sqrt{1+x^2}-x)}{(\sqrt{1+x^2})^2-x^2} \, dx=\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} x\sqrt{1+x^2}\cos x \, dx-\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} x^2 \cos x \, dx =-2\int_{0}^{\frac{\pi}{6}} x^2 \cos x \, dx$ as the function $\displaystyle x\sqrt{1+x^2}\cos x$ is an odd function and $\displaystyle x^2 \cos x$ is an even function. Thus, $I=-2\int_{0}^{\frac{\pi}{6}} x^2 \cos x \, dx=-2\left .\left( x^2 \sin x+2x \cos x-2 \sin x\right)\right|_{0}^{\frac{\pi}{6}}=\boxed{2-\frac{\pi}{\sqrt{3}}-\frac{\pi^2}{36}}.$