Question

$\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx=?$

• A. $-\frac{1}{2}\sin 2x+c$
• B. $\frac{1}{2}\sin 2x+c$
• C. $\frac{1}{2}\sin x+c$
• D. $-\frac{1}{2}\sin x+c$

Answer 1

The correct option is A

$-\frac{1}{2}\sin 2x+c$

Explanation for the correct option:

Evaluating the given integral:

$\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx$

$$\begin{gathered} =\int \frac{{\left({\sin }^{4}x\right)}^2-{\left({\cos }^{4}x\right)}^2}{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x}dx\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right] \\ =\int \frac{\left({\sin }^{4}x+{\cos }^{4}x\right)\left({\sin }^{4}x-{\cos }^{4}x\right)}{\left({\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x-2{\sin }^{2}x{\cos }^{2}x\right)}dx\left[\because (a^2-b^2)=(a-b)(a+b)\&{(a+b)}^2=a^2+b^2+2ab\right] \\ =\int \frac{\left({\sin }^{4}x+{\cos }^{4}x\right)\left({\sin }^{4}x-{\cos }^{4}x\right)}{\left({\sin }^{4}x+{\cos }^{4}x\right)}dx \\ =\int \left({\sin }^{4}x-{\cos }^{4}x\right)dx \\ =\int \left({\sin }^{2}x-{\cos }^{2}x\right)\left({\sin }^{2}x+{\cos }^{2}x\right)dx\left[\because (a^2-b^2)=(a-b)(a+b)\right] \\ =\int \left({\sin }^{2}x-{\cos }^{2}x\right)dx \\ =\int \left(1-{\cos }^{2}x-{\cos }^{2}x\right)dx\left[\because {\sin }^{2}x=1-{\cos }^{2}x\right] \\ =\int \left(1-2{\cos }^{2}x\right)dx \\ =-\int \left(-1+2{\cos }^{2}x\right)dx \\ =-\int \left(2{\cos }^{2}x-1\right)dx \\ =-\int \cos \left(2x\right)dx\left[\because \cos \left(2x\right)=2{\cos }^{2}x-1\right] \\ =-\frac{\sin \left(2x\right)}{2}+C\left[C\text{is an integrating constant}\right] \end{gathered}$$

Therefore, $\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x\cos 2x}dx=$$-\frac{1}{2}\sin 2x+c$

Hence, the correct answer is option (B).