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| cos\x + 1/ dx = | cos\1 + x / dx
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$$-{{\sqrt{\pi}\,\left(\left(\left(\sqrt{2}\,i+\sqrt{2}\right)\,\sin
1+\left(\sqrt{2}\,i-\sqrt{2}\right)\,\cos 1\right)\,\mathrm{erf}
\left({{\sqrt{2}\,i+\sqrt{2}}\over{2}}\right)+\left(\left(\sqrt{2}\,
i-\sqrt{2}\right)\,\sin 1+\left(\sqrt{2}\,i+\sqrt{2}\right)\,\cos 1
\right)\,\mathrm{erf}\left({{\sqrt{2}\,i-\sqrt{2}}\over{2}}\right)+
\left(\left(\sqrt{2}-\sqrt{2}\,i\right)\,\sin 1+\left(-\sqrt{2}\,i-
\sqrt{2}\right)\,\cos 1\right)\,\mathrm{erf}\left(\sqrt{-i}\right)+
\left(\left(\sqrt{2}\,i+\sqrt{2}\right)\,\sin 1+\left(\sqrt{2}\,i-
\sqrt{2}\right)\,\cos 1\right)\,\mathrm{erf}\left(\left(-1\right)^{
{{1}\over{4}}}\right)\right)}\over{16}}$$
Ответ (Неопределённый)
[src]$$-{{\sqrt{\pi}\,\left(\left(\left(\sqrt{2}\,i+\sqrt{2}\right)\,\sin
1+\left(\sqrt{2}\,i-\sqrt{2}\right)\,\cos 1\right)\,\mathrm{erf}
\left({{\left(\sqrt{2}\,i+\sqrt{2}\right)\,x}\over{2}}\right)+\left(
\left(\sqrt{2}\,i-\sqrt{2}\right)\,\sin 1+\left(\sqrt{2}\,i+\sqrt{2}
\right)\,\cos 1\right)\,\mathrm{erf}\left({{\left(\sqrt{2}\,i-\sqrt{
2}\right)\,x}\over{2}}\right)+\left(\left(\sqrt{2}-\sqrt{2}\,i
\right)\,\sin 1+\left(-\sqrt{2}\,i-\sqrt{2}\right)\,\cos 1\right)\,
\mathrm{erf}\left(\sqrt{-i}\,x\right)+\left(\left(\sqrt{2}\,i+\sqrt{
2}\right)\,\sin 1+\left(\sqrt{2}\,i-\sqrt{2}\right)\,\cos 1\right)\,
\mathrm{erf}\left(\left(-1\right)^{{{1}\over{4}}}\,x\right)\right)
}\over{16}}$$