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## Indefinite integral (01)

Calculus (Integrals, Series)
Grigorios Kostakos
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### Indefinite integral (01)

Prove that
$$\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}}\,dx}=\frac{3}{5}\,\arctan({\sin{x}-2\,\cos{x}})+\frac{\sqrt{6}}{60}\,\ln\Bigl|{\tfrac{\sqrt{6}+2\,\sin{x}+\cos{x}}{\sqrt{6}-2\,\sin{x}-\cos{x}}}\Bigr|$$
Grigorios Kostakos
Grigorios Kostakos
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Articles: 0
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Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

### Re: Indefinite integral (01)

$I=\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}}\,dx}=\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-4\,\sin{x}\,\cos{x}+2\,\sin^2{x}}\,dx}.$

Substituting $t=\tan{x}$, we have $\sin{x}=\displaystyle\frac{t}{\sqrt{1+t^2}}$, $\cos{x}=\displaystyle\frac{1}{\sqrt{1+t^2}}$ and $dx=\displaystyle\frac{1}{1+t^2}\,dt$.

$I=\displaystyle\int{\frac{\frac{1}{\sqrt{1+t^2}}+\frac{t}{\sqrt{1+t^2}}}{5\,\frac{1}{1+t^2}-4\,\frac{t}{1+t^2}+2\,\frac{t^2}{1+t^2}}\,\frac{1}{1+t^2}\,dt}=\int{\frac{1+t}{({5-4t+2t^2})\,\sqrt{1+t^2}}\,dt}=$

$\displaystyle\int{\frac{\frac{3}{5}({2t+1})-\frac{\sqrt{6}}{30}\,\sqrt{6}\,({t-2})}{({2t^2-4t+5})\,\sqrt{1+t^2}}\,dt}=$

$\displaystyle\frac{3}{5}\int{\frac{2t+1}{\left({({t-2})^2+t^2+1}\right)\sqrt{1+t^2}}\,dt}-\frac{\sqrt{6}}{30}\int{\frac{\sqrt{6}\,({t-2})}{\bigl({\bigl({\sqrt{6}\,\sqrt{1+t^2}\,}\bigl)^2-({2t+1})^2}\bigl)\sqrt{1+t^2}}\,dt}=$

$\displaystyle\frac{3}{5}\int{\frac{2t+1}{\Bigl({\bigl({\frac{t-2}{\sqrt{1+t^2}}}\bigr)^2+1}\Bigr)\,({1+t^2})^{\frac{3}{2}}}\,dt}-\frac{\sqrt{6}}{30}\int{\frac{\sqrt{6}\,({t-2})}{\Bigl({\bigl({\frac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\bigr)^2-1}\Bigr)({2t+1})^2\sqrt{1+t^2}}\,dt}=$

$\displaystyle\frac{3}{5}\int{\frac{1}{\bigl({\frac{t-2}{\sqrt{1+t^2}}}\bigr)^2+1}\,d\bigl({\tfrac{t-2}{\sqrt{1+t^2}}}\bigr)}-\frac{\sqrt{6}}{30}\int{\frac{1}{\Bigl({\frac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\Bigr)^2-1}\,d\Bigl({\tfrac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\Bigr)}=$

$\displaystyle\frac{3}{5}\int{\frac{1}{\bigl({\tfrac{t-2}{\sqrt{1+t^2}}}\bigr)^2+1}\,d\bigl({\tfrac{t-2}{\sqrt{1+t^2}}}\bigr)}-\frac{\sqrt{6}}{30}\int{\frac{1}{\Bigl({\frac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}-1}\Bigr)\,\Bigl({\frac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}+1}\Bigr)}\,d\Bigl({\tfrac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}}\Bigr)}=$

$\displaystyle\frac{3}{5}\,\arctan\bigl({\tfrac{t-2}{\sqrt{1+t^2}}}\bigr)-\frac{\sqrt{6}}{60}\,\Bigl({\ln\bigl|{\tfrac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}-1}\bigr|-\ln\bigl|{\tfrac{\sqrt{6}\,\sqrt{1+t^2}}{2t+1}+1}\bigr|}\Bigr)+c=$

$\displaystyle\frac{3}{5}\,\arctan\bigl({\tfrac{t-2}{\sqrt{1+t^2}}}\bigr)+\frac{\sqrt{6}}{60}\,\ln\Bigl|{\tfrac{\sqrt{6}\,\sqrt{1+t^2}+2t+1}{\sqrt{6}\,\sqrt{1+t^2}-2t-1}}\Bigr|+c=$

$\displaystyle\frac{3}{5}\,\arctan\bigl({\tfrac{t}{\sqrt{1+t^2}}-2\,\tfrac{1}{\sqrt{1+t^2}}}\bigr)+\frac{\sqrt{6}}{60}\,\ln\Bigg|{\tfrac{\frac{\sqrt{6}\,\sqrt{1+t^2}+2t+1}{\sqrt{1+t^2}}}{\frac{\sqrt{6}\,\sqrt{1+t^2}-2t-1}{\sqrt{1+t^2}}}}\Bigg|+c=$

$\displaystyle\frac{3}{5}\,\arctan\bigl({\tfrac{t}{\sqrt{1+t^2}}-2\,\tfrac{1}{\sqrt{1+t^2}}}\bigr)+\frac{\sqrt{6}}{60}\,\ln\bigg|{\tfrac{\sqrt{6}+2\,\frac{t}{\sqrt{1+t^2}}+\frac{1}{\sqrt{1+t^2}}}{\sqrt{6}-2\,\frac{t}{\sqrt{1+t^2}}-\frac{1}{\sqrt{1+t^2}}}}\bigg|+c\,\stackrel{t=\tan{x}}{=\!=\!=\!=}$

$\displaystyle\frac{3}{5}\,\arctan({\sin{x}-2\,\cos{x}})+\frac{\sqrt{6}}{60}\,\ln\Bigl|{\tfrac{\sqrt{6}+2\,\sin{x}+\cos{x}}{\sqrt{6}-2\,\sin{x}-\cos{x}}}\Bigr|+c.\quad\square$
Grigorios Kostakos
Articles: 1
Posts: 40
Joined: Mon Oct 26, 2015 12:27 pm

### Re: Indefinite integral (01)

A second solution given by Prof. S. Ntougias:

It is known that
\begin{align*}