Question
Evaluate the following integrals:$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
✓
Answer
$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Let $\text{t}=\cos^2\text{x}\rightarrow-\text{dt}=2\cos\text{x}\sin\text{x}\text{ dx}$
$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
$=\int\frac{-1}{\sqrt{\text{t}^2-(1-\text{t})+2}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+1}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2+\frac{3}{4}}}\text{ dt}$
$=-\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\text{t}^2+\text{t}+1}\Big|$
$=-\log\Big|\Big(\cos^2\text{x}+\frac{1}{2}\Big)+\sqrt{\cos^4\text{x}+\cos^2\text{x}+1}\Big|+\text{C}$
Let $\text{t}=\cos^2\text{x}\rightarrow-\text{dt}=2\cos\text{x}\sin\text{x}\text{ dx}$
$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
$=\int\frac{-1}{\sqrt{\text{t}^2-(1-\text{t})+2}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+1}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2+\frac{3}{4}}}\text{ dt}$
$=-\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\text{t}^2+\text{t}+1}\Big|$
$=-\log\Big|\Big(\cos^2\text{x}+\frac{1}{2}\Big)+\sqrt{\cos^4\text{x}+\cos^2\text{x}+1}\Big|+\text{C}$
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