Question

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

Answer 1

\[\int\frac{\text{ sin }\left( 2 x \right) dx}{\sqrt{\cos^4 x - \sin^2 x + 2}}\]
` ⇒ ∫ {2 sin x cos x  dx}/{\sqrt{cos^4 x - \left( 1 - \cos^2 x \right) + 2}}`
\[ \Rightarrow \int\frac{2 \sin x \cos x}{\sqrt{\cos^4 x + \cos^2 x + 1}}\]
\[\text{ Let } \cos^2 x = t\]
\[ \Rightarrow 2 \cos x \times - \text{ sin x dx } = dt\]
\[\text{ sin } \left( 2x \right) dx = - dt\]
\[Now, \int\frac{\sin \left( 2 x \right) dx}{\sqrt{\cos^4 x - \sin^2 x + 2}}\]
\[ = \int\frac{- dt}{\sqrt{t^2 + t + 1}}\]
\[ = \int\frac{- dt}{\sqrt{t^2 + t + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 1}}\]
\[ = - \int\frac{dt}{\sqrt{\left( t + \frac{1}{2} \right)^2 + \frac{3}{4}}}\]
\[ = - \int\frac{dt}{\sqrt{\left( t + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}}\]
\[ = - \text{ log }\left| t + \frac{1}{2} + \sqrt{\left( t + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right| + C\]
\[ = - \text{ log }\left| t + \frac{1}{2} + \sqrt{t^2 + t + 1} \right| + C\]
\[ = - \text{ log }\left| \cos^2 x + \frac{1}{2} + \sqrt{\cos^4 x + \cos^2 x + 1} \right| + C\]