Integral exercise asked by Hiềntài Là Nguyênkhíquốctế in the pretty math group

Exercise:

Integrate, \(\int{\frac{\sin x}{\sqrt{\sin 2x+2}}dx}\)

Solution: we have $\sin 2x=2\sin x\cos x+2=2\sin x\cos x+{{\cos }^{2}}x+{{\sin }^{2}}x+1$

$\Rightarrow \sin 2x+2={{\left( \cos x+\sin x \right)}^{2}}+1$

So $\int{\frac{\sin x}{\sqrt{\sin 2x+1}}dx}=\frac{1}{2}\int{\frac{2\sin x}{\sqrt{{{\left( \cos x+\sin x \right)}^{2}}+1}}dx=\frac{1}{2}\int{\frac{\sin x+\sin x-\cos x+\cos x}{\sqrt{{{\left( \cos x+\sin x \right)}^{2}}+1}}dx}}$

$=\frac{1}{2}\int{\frac{-\left( \cos x-\sin x \right)}{\sqrt{{{\left( \cos x+\sin x \right)}^{2}}+1}}dx+\frac{1}{2}}\int{\frac{\sin x+\cos x}{\sqrt{{{\left( \cos x+\sin x \right)}^{2}}+1}}dx}$

Let $u=\cos x+\sin x\Leftrightarrow du=\left( \cos x-\sin x \right)dx$

$\Rightarrow -\int{\frac{\cos x-\sin x}{\sqrt{{{\left( \cos x+\sin x \right)}^{2}}+1}}dx}=-\int{\frac{du}{\sqrt{{{u}^{2}}+1}}}$

Take $u=\tan w\Leftrightarrow du={{\sec }^{2}}wdw$

So $\int{\frac{du}{\sqrt{{{u}^{2}}+1}}=\int{\frac{{{\sec }^{2}}w}{\sqrt{{{\tan }^{2}}w+1}}dw}=\int{\sec w\,dw}}=\int{\frac{\sec w\left( \sec w+\tan w \right)}{\sec w+\tan w}dw}=\ln \left| \sec w+\tan w \right|+c$

$=\ln \left| \sec w+u \right|+c$ but ${{\tan }^{2}}w+1={{\sec }^{2}}w\Leftrightarrow \sec w=\sqrt{{{\tan }^{2}}w+1}=\sqrt{{{u}^{2}}+1}$

Hence $\int{\frac{du}{\sqrt{{{u}^{2}}+1}}=\ln \left| \sqrt{{{u}^{2}}+1}+u \right|+c}$

Thus $-\frac{1}{2}\int{\frac{\cos x-\sin x}{\sqrt{{{\left( \cos x+\sin x \right)}^{2}}+1}}dx=-\frac{1}{2}\ln \left| \sqrt{{{\left( \cos x+\sin x \right)}^{2}}+1}+\cos x+\sin x \right|+c}$   (*1*)

$\int{\frac{\sin x+\cos x}{\sqrt{{{\left( \sin x+\cos x \right)}^{2}}+1}}dx}=-\int{\frac{-\sin x-\cos x}{\sqrt{2\cos x\sin x-1+3}}dx=-\int{\frac{-\sin x-\cos x}{\sqrt{3-{{\left( \cos x-\sin x \right)}^{2}}}}dx}}$

Take $y=\cos x-\sin x\Leftrightarrow dy=\left( -\sin x-\cos x \right)dx$

So $-\int{\frac{-\sin x-\cos x}{\sqrt{3-{{\left( \cos x-\sin x \right)}^{2}}}}dx}=-\int{\frac{dy}{\sqrt{3-{{y}^{2}}}}}$

Put $y=\sqrt{3}\sin t\Leftrightarrow dy=\sqrt{3}\cos t\,dt$

So $-\int{\frac{dy}{\sqrt{3-{{y}^{2}}}}=-\int{\frac{\sqrt{3}\cos t}{\sqrt{3-3{{\sin }^{2}}t}}dt=-\sqrt{3}\int{\frac{\cos t}{\sqrt{3\left( 1-{{\sin }^{2}}t \right)}}dt}=-\int{\frac{\cos t}{\cos t}dt}=-t+c}}$

But $y=\sqrt{3}\sin t\Rightarrow t=\arcsin \left( \frac{y}{\sqrt{3}} \right)$

Hence $-\int{\frac{-\sin x-\cos x}{\sqrt{3-{{\left( \cos x-\sin x \right)}^{2}}}}dx=-\arcsin \left( \frac{\cos x-\sin x}{\sqrt{3}} \right)+c}$   .  (*2*)

Thus

$\int{\frac{\sin x}{\sqrt{\sin 2x+2}}dx}=-\frac{1}{2}\ln \left| \sqrt{\sin 2x+2}+\cos x+\sin x \right|-\frac{1}{2}\arcsin \left( \frac{\cos x-\sin x}{\sqrt{3}} \right)+c$