Wolframalpha step-by-step of $\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$

Question

I wonder, where the minus sign goes after the first $u$-substitution of integral $\displaystyle\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$?

Answer 1

Integrate

$$\int \frac{\cos x}{\sqrt{2+\cos(2x)}}dx$$

Known Identity

$$\cos(2x)=1-2\sin^2(x)$$

Replacing

$$\int \frac{\cos x}{\sqrt{3-2\sin^2(x)}}dx$$

Let $u = \sin(x)$, $du = \cos(x)dx$

Substituting

$$\int \frac{u}{\sqrt{3-2u^2}}du$$

Let $z = \sqrt{3-2u^2}$

$$dz = \frac{1}{2}\frac{-4u}{\sqrt{3-2u^2}}du$$ $$dz = -\frac{2u}{\sqrt{3-2u^2}}du$$

Substituting $$-\int \frac{1}{2}dz$$ $$=-\frac{z}{2} + C$$ Substituting back $$=- \frac{1}{2} \sqrt{- 2 u^{2} + 3}+ C$$

Substituting back $u = \sin(x)$ $$=-\frac{1}{2}\sqrt{-2\sin^2(x)+3}+ C$$ Simplifying $$=-\frac{1}{2}\sqrt{2(1-\sin^2(x))+1}+ C$$ $$=-\frac{1}{2}\sqrt{2\cos^2(x)+1}+ C$$ $$=-\frac{1}{2}\sqrt{1+2\cos^2(x)}+ C$$

Answer 2

Rewrite $2+\cos 2x$ as $3-2\sin^2 x$ and let $u=\sin x$. We end up at $$\int \frac{du}{\sqrt{3-2u^2}}.$$ Let $\sqrt{2}u=\sqrt{3}{v}$. Then $du=\frac{\sqrt{3}}{\sqrt{2}}\,dv$, and we end up with $$\int\frac{1}{\sqrt{2}}\frac{dv}{\sqrt{1-v^2}}.$$ Thus our integral is $\frac{1}{\sqrt{2}}\arcsin v+C$. Replace $v$ by $\frac{\sqrt{2}\sin x}{\sqrt{3}}$.