Question

$\displaystyle \int \frac{sin^{8} \text{x} - cos^{8} \text{x}}{1 - 2 sin^{2} \text{x} cos^{2} \text{x}} \text{dx}$ is equal to (where $C$ is an arbitrary constant)

• A. $\frac{1}{2} sin2x + \text{C}$
• B. $- \frac{1}{2} sin2x + \text{C}$
• C. $- \frac{1}{2} sinx + \text{C}$
• D. $- sin^{2} \text{x} + \text{C}$

Answer 1

Answer: (B)

$\displaystyle \int \frac{\text{sin}^{8} \text{x} - \text{cos}^{8} \text{x}}{1 - \text{2sin}^{2} \text{xcos}^{2} \text{x}}\text{dx}$
$=\displaystyle \int \frac{\left(\left(\text{sin}\right)^{4} \text{x} - \left(\text{cos}\right)^{4} \text{x}\right) \left(\left(\text{sin}\right)^{4} \left(\text{x + cos}\right)^{4} \text{x}\right)}{1 - \left(\text{2sin}\right)^{2} \left(\text{xcos}\right)^{2} \text{x}}\text{dx}$
$=\displaystyle \int \frac{\left(\left(\text{sin}\right)^{2} \text{x} - \left(\text{cos}\right)^{2} \text{x}\right) \left(\left(\text{sin}\right)^{4} \left(\text{x + cos}\right)^{4} \text{x}\right)}{1 - \left(\text{2 sin}\right)^{2} \left(\text{xcos}\right)^{2} \text{x}}\text{dx}$
$=\displaystyle \int -\text{cos2x dx}=-\frac{1}{2}\text{sin2x + C}$