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I have just done these, but I don't know what to do next......

\begin{align} \int \frac{\sin x}{x+x\cos^2 x}\,\mathrm{d}x & = \int \frac{1}{x}\cdot\frac{\sin x}{1+\cos^2 x}\,\mathrm{d}x\\ &=\int \frac{1}{x}\,\mathrm{d}(\arctan(-\cos x))\\\ &=\frac{1}{x} \arctan(-\cos x)-\int -\frac{1}{x^2} \arctan(-\cos x)\,\mathrm{d}x \\ &=\frac{1}{x} \arctan(-\cos x)-\int \frac{\arctan(\cos x)}{x^2}\,\mathrm{d}x \end{align}

NoName
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Haoxiang Yang
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1 Answers1

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Apply Lobachevsky's formula: if $f\in\mathcal{C}(\mathbb{R}_+)$ meets $f(\pi\pm x)=f(x)$, then $$\int_0^\infty f(x)\,\frac{\sin x}{x}\,dx=\int_0^{\pi/2}f(x)\,dx.$$ The given integral is then equal to $$\int_0^{\pi/2}\frac{dx}{1+\cos^2 x}\ \underset{t=\tan x}{\phantom{\big[}=\phantom{\big]}}\ \int_0^\infty\frac{dt}{2+t^2}=\frac\pi{2\sqrt2}.$$

metamorphy
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