Question

If $\int \frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}dx$ = a sin 2x + C, then a =

(a) −1/2
(b) 1/2
(c) −1
(d) 1

Answer 1

(a) −1/2


$$\begin{gathered} \mathrm{If}\int \left(\frac{{\sin }^{8}x-{\cos }^{8}x}{1-2{\sin }^{2}x{\cos }^{2}x}\right)dx=a\sin 2x+C\mathit{}\mathit{}\mathit{}....\left(1\right) \\ \mathrm{Considering}\mathrm{LHS}\mathrm{of}\mathrm{eq}\left(1\right) \\ \Rightarrow \int \frac{\left({\sin }^{4}x-{\cos }^{4}x\right)\left({\sin }^{4}x+{\cos }^{4}x\right)}{\left(1-2{\sin }^{2}x{\cos }^{2}x\right)} \\ \Rightarrow \int \frac{\left({\sin }^{2}x-{\cos }^{2}x\right)\left({\sin }^{2}x+{\cos }^{2}x\right)·\left({\sin }^{4}x+{\cos }^{4}x\right)dx}{\left\{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x\right\}} \\ \Rightarrow \int \frac{\left({\sin }^{2}x-{\cos }^{2}x\right)·\left({\sin }^{4}x+{\cos }^{4}x\right)dx}{\left({\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x-2{\sin }^{2}x{\cos }^{2}x\right)} \\ \Rightarrow -\int \frac{\left({\cos }^{2}x-{\sin }^{2}x\right)\times \left({\sin }^{4}x+{\cos }^{4}x\right)dx}{\left({\sin }^{4}x+{\cos }^{4}x\right)} \\ \Rightarrow -\int \cos \left(2x\right)dx\left(\because {\cos }^{2}x-{\sin }^{2}x=\cos 2x\right)...\left(2\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{RHS}\mathrm{of}\mathrm{eq}\left(1\right)\mathrm{with}\mathrm{eq}\left(2\right)\mathrm{we}\mathrm{get}, \\ a=-\frac{1}{2} \end{gathered}$$