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Indefinite integral (01)

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

#1

Post by Grigorios Kostakos » Wed Nov 11, 2015 3:46 am

Prove that
$$\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}}\,dx}=\frac{3}{5}\,\arctan({\sin{x}-2\,\cos{x}})+\frac{\sqrt{6}}{60}\,\ln\Bigl|{\tfrac{\sqrt{6}+2\,\sin{x}+\cos{x}}{\sqrt{6}-2\,\sin{x}-\cos{x}}}\Bigr|$$
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Re: Indefinite integral (01)

#2

Post by Grigorios Kostakos » Wed Nov 11, 2015 3:47 am

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

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 \(dx=\displaystyle\frac{1}{1+t^2}\,dt\).

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

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

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

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

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

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

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

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

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

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

\(\displaystyle\frac{3}{5}\,\arctan({\sin{x}-2\,\cos{x}})+\frac{\sqrt{6}}{60}\,\ln\Bigl|{\tfrac{\sqrt{6}+2\,\sin{x}+\cos{x}}{\sqrt{6}-2\,\sin{x}-\cos{x}}}\Bigr|+c.\quad\square\)
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Re: Indefinite integral (01)

#3

Post by admin » Wed Nov 11, 2015 3:49 am

A second solution given by Prof. S. Ntougias:

It is known that
\begin{align*}
& \cos{x}+\sin{x}=\frac{3}{5}\left({\cos{x}+2\,\sin{x}}\right)+\frac{1}{5}\left({2\,\cos{x}-\sin{x}}\right)\quad{\text{ and}}\\
& 5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}=1+\left({\sin{x}-2\,\cos{x}}\right)^2=6-\left({2\,\sin{x}+\cos{x}}\right)^2\,.
\end{align*}
So
\begin{align*}
&\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}}\,dx}=\\
&\displaystyle\frac{3}{5}\int{\frac{d\left({\sin{x}-2\,\cos{x}}\right)}{1+\left({\sin{x}-2\,\cos{x}}\right)^2}}+\frac{1}{5}\int{\frac{d\left({2\,\sin{x}+\cos{x}}\right)}{6-\left({2\,\sin{x}+\cos{x}}\right)^2}}=\\
& \frac{3}{5}\int{\frac{d\left({\sin{x}-2\,\cos{x}}\right)}{1+\left({\sin{x}-2\,\cos{x}}\right)^2}}-\frac{\sqrt{6}}{60}\int{\frac{d\left({\sqrt{6}-2\,\sin{x}-\cos{x}}\right)}{\sqrt{6}-2\,\sin{x}-\cos{x}}}+\frac{\sqrt{6}}{60}\int{\frac{d\left({\sqrt{6}+2\,\sin{x}+\cos{x}}\right)}{\sqrt{6}+2\,\sin{x}+\cos{x}}}=\\
&\displaystyle\frac{3}{5}\,\arctan\left({\sin{x}-2\,\cos{x}}\right)+\frac{\sqrt{6}}{60}\,\ln\left|{\tfrac{\sqrt{6}+2\,\sin{x}+\cos{x}}{\sqrt{6}-2\,\sin{x}-\cos{x}}}\right|+c\,.\quad\square
\end{align*}
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