Indefinite integral (05)

[Grigorios Kostakos]

Find \(\displaystyle\int{\frac{\sin{x}+\cos{x}}{\sqrt{\sin(2x)}}\,dx}\,\)

assuming that \(\sin{x}>0\) and \(\cos{x}>0\).

[Papapetros Vaggelis]

For all \(\displaystyle{x\in\mathbb{R}}\)

\(\displaystyle{(\sin x-\cos x)^2=1-\sin (2x)\Rightarrow \sin (2x)=1-(\sin x-\cos x)^2}\).

Because,

\(\displaystyle{\sin x>0}\) and \(\cos x>0\),

for all \(\displaystyle{x\in\left(k\pi,k\pi+\frac{\pi}{2}\right)}\), \(\displaystyle{k=2m,m\in\mathbb{N}}\)

\(\displaystyle{\int \frac{\sin x+\cos x}{\sqrt{\sin (2x)}}dx=}\)

\(\displaystyle{=\int \frac{(\sin x-\cos x)'}{\sqrt{1-(\sin x-\cos x)^2}}dx}\)

\(\displaystyle{=\arcsin (\sin x-\cos x)+c,c\in\mathbb{R}}\).

The last expression is well defined because,

\(\displaystyle{\left(0<\sin x\leq 1\right)\ \land \left(-1\leq -\cos x<0\right)\Rightarrow -1<\sin x-\cos x<1}\)

[Grigorios Kostakos]

A 2nd solution:

\begin{align*}
\displaystyle\int{\frac{\sin{x}+\cos{x}}{\sqrt{\sin(2x)}}\,dx}&=\int{\frac{\sin{x}+\cos{x}}{\sqrt{2\,\sin{x}\,\cos{x}}}\,dx}\\
&=\frac{1}{\sqrt{2}}\int{\frac{{\sin{x}}}{\sqrt{\sin{x}}\,\sqrt{\cos{x}}}+\frac{{\cos{x}}}{\sqrt{\sin{x}}\,\sqrt{\cos{x}}}\,dx}\\
&=\frac{1}{\sqrt{2}}\int{\sqrt{\frac{\sin{x}}{\cos{x}}}+\frac{1}{\sqrt{\frac{\sin{x}}{\cos{x}}}}\,dx}\\
&=\frac{1}{\sqrt{2}}\int{\sqrt{\tan{x}}+\frac{1}{\sqrt{\tan{x}}}\,dx}\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{{\begin{subarray}{c}
{t\,=\,\sqrt{\tan{x}}} \\
{\frac{2t}{1+t^4}\,dt\,=\,dx}
\end{subarray}}}\,\frac{1}{\sqrt{2}}\int{\Bigl({t+\frac{1}{t}}\Bigr)\,\frac{2t}{1+t^4}\,dt}\\
&=\frac{1}{\sqrt{2}}\int{\frac{2t^2+2}{t^4+1}\,dt}\\
&=\frac{1}{\sqrt{2}}\int{\frac{2t^2+2}{t^4+2t^2+1-2t^2}\,dt}\\
&=\frac{1}{\sqrt{2}}\int{\frac{2t^2+2}{\bigl({t^2+1}\bigr)^2-\bigl({t\sqrt{2}\,}\bigr)^2}\,dt}\\
&=\frac{1}{\sqrt{2}}\int{\frac{2t^2+2}{\bigl({t^2+1-t\sqrt{2}\,}\bigr)\,\bigl({t^2+1+t\sqrt{2}\,}\bigr)}\,dt}\\
&=\frac{1}{\sqrt{2}}\int{\frac{1}{t^2+1-t\sqrt{2}}+\frac{1}{t^2+1+t\sqrt{2}}\,dt}\\
&=\frac{1}{\sqrt{2}}\int{\frac{1}{t^2+1-t\sqrt{2}}\,dt}+\frac{1}{\sqrt{2}}\int{\frac{1}{t^2+1+t\sqrt{2}}\,dt}\\
&=\frac{1}{\sqrt{2}}\int{\frac{1}{t^2+1-t\sqrt{2}}+\frac{1}{t^2+1+t\sqrt{2}}\,dt}\\
&=\frac{1}{\sqrt{2}}\int{\frac{2}{\bigl({t\sqrt{2}-1}\bigr)^2+1}\,dt}+\frac{1}{\sqrt{2}}\int{\frac{2}{\bigl({t\sqrt{2}+1}\bigr)^2+1}\,dt}\\
&=\int{\frac{1}{\bigl({t\sqrt{2}-1}\bigr)^2+1}\,d\bigl({t\sqrt{2}-1}\bigr)}+\int{\frac{1}{\bigl({t\sqrt{2}+1}\bigr)^2+1}\,d\bigl({t\sqrt{2}+1}\bigr)}\\
&=\arctan\bigl({t\sqrt{2}-1}\bigr)+\arctan\bigl({t\sqrt{2}+1}\bigr)+c\\
&=\arctan\Bigl({\tfrac{t\sqrt{2}-1+t\sqrt{2}+1}{1-({t\sqrt{2}-1})\,({t\sqrt{2}+1})}}\Bigr)+c\\
&=\arctan\bigl({\tfrac{t\sqrt{2}}{1-t^2}}\bigr)+c\\
&\mathop{=\!=\!=\!=\!=}\limits^{t\,=\,\sqrt{\tan{x}}}\,\arctan\bigl({\tfrac{\sqrt{2\tan{x}}}{1-\tan{x}}\,}\bigr)+c\,.\qquad\square
\end{align*}