Question

$\int \frac{si{n}^{8}x-co{s}^{8}x}{1-2si{n}^{2}xco{s}^{2}x}dx$ is equal to

• A. $\frac{1}{2}sin2x+c$
• B. $-\frac{1}{2}sin2x+c$
• C. $-\frac{1}{2}sinx+c$
• D. $-si{n}^{2}x+c$

Answer 1

The correct option is B $-\frac{1}{2}sin2x+c$

$I=\int \frac{si{n}^{8}x-co{s}^{8}x}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{(si{n}^{4}x-co{s}^{4}x)(si{n}^{4}x+co{s}^{4}x)}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{(si{n}^{2}x-co{s}^{2}x)(si{n}^{2}x+co{s}^{2}x)(si{n}^{4}x+co{s}^{4}x)}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{1.(si{n}^{2}x-co{s}^{2}x)\left[\right(si{n}^{2}x+co{s}^{2}x{)}^2-2si{n}^{2}co{s}^{2}x]}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{(si{n}^{2}x-co{s}^{2}x)(1-2si{n}^{2}xco{s}^{2}x)}{1-2si{n}^{2}xco{s}^{2}x}dx=-\int cos2xdx=-\frac{1}{2}sin2x+c$