Question

$\int \frac{\sin 2x}{\sqrt{{\cos }^{4}x-{\sin }^{2}x+2}}dx$

Answer 1

$$\begin{gathered} \int \frac{\sin \left(2x\right)dx}{\sqrt{{\cos }^{4}x-{\sin }^{2}x+2}} \\ \Rightarrow \int \frac{2\sin x\cos xdx}{\sqrt{{\cos }^{4}x-\left(1-{\cos }^{2}x\right)+2}} \\ \Rightarrow \int \frac{2\sin x\cos x}{\sqrt{{\cos }^{4}x+{\cos }^{2}x+1}} \\ \mathrm{Let}{\cos }^{2}x=t \\ \Rightarrow 2\cos x\times -\sin xdx=dt \\ \sin \left(2x\right)dx=-dt \\ \mathrm{Now},\int \frac{\sin \left(2x\right)dx}{\sqrt{{\cos }^{4}x-{\sin }^{2}x+2}} \\ =\int \frac{-dt}{\sqrt{t^2+t+1}} \\ =\int \frac{-dt}{\sqrt{t^2+t+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2+1}} \\ =-\int \frac{dt}{\sqrt{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+{\displaystyle \frac{3}{4}}}} \\ =-\int \frac{dt}{\sqrt{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}} \\ =-\log \left|t+\frac{1}{2}+\sqrt{{\left(t+\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\right|+C \\ =-\log \left|t+\frac{1}{2}+\sqrt{t^2+t+1}\right|+C \\ =-\log \left|{\cos }^{2}x+\frac{1}{2}+\sqrt{{\cos }^{4}x+{\cos }^{2}x+1}\right|+C \end{gathered}$$