Question

If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]

Options

• -1/2

• 1/2

• -1

• 1

Answer 1

`−1/2`

 

\[\text{If }\int\left( \frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} \right)dx = a \sin 2x + C ..............(1)\]

\[\text{Considering LHS of eq. (1)}\]

\[ \Rightarrow \int\frac{\left( \sin^4 x - \cos^4 x \right) \left( \sin^4 x + \cos^4 x \right)}{\left( 1 - 2 \sin^2 x \cos^2 x \right)}\]

\[ \Rightarrow \int\frac{\left( \sin^2 x - \cos^2 x \right) \left( \sin^2 x + \cos^2 x \right) \cdot \left( \sin^4 x + \cos^4 x \right) dx}{\left\{ \left( \sin^2 x + \cos^2 x \right)^2 - 2 \sin^2 x \cos^2 x \right\}}\]

\[ \Rightarrow \int\frac{\left( \sin^2 x - \cos^2 x \right) \cdot \left( \sin^4 x + \cos^4 x \right)dx}{\left( \sin^4 x + \cos^4 x + 2 \sin^2 x \cos^2 x - 2 \sin^2 x \cos^2 x \right)}\]

\[ \Rightarrow - \int\frac{\left( \cos^2 x - \sin^2 x \right) \times \left( \sin^4 x + \cos^4 x \right) dx}{\left( \sin^4 x + \cos^4 x \right)}\]

\[ \Rightarrow - \int\cos \left( 2x \right) dx ..............\left( \because \cos^2 x - \sin^2 x = \cos 2x \right) .............(2)\]

\[\text{Comparing the RHS of eq. (1) with eq. (2) we get,} \]

\[a = - \frac{1}{2}\]