Question

$\int \frac{si{n}^8\text{x}-co{s}^8\text{x}}{1-2si{n}^2\text{x}co{s}^2\text{x}}\text{dx}$ 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. $-si{n}^2\text{x}+\text{C}$

Answer 1

Answer: (B)

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