If $\int \frac{{\mathrm{x}}^7}{{\left(1-{\mathrm{x}}^2\right)}^5}\mathrm{dx}=\frac{1}{\mathrm{\lambda }}\frac{{\mathrm{x}}^8}{{\left(1-{\mathrm{x}}^2\right)}^4}+\mathrm{C}$ then $\lambda$  is equal to

we have,

$\begin{array}{l}\mathrm{I}=\int \frac{{\mathrm{x}}^7}{{\left(1-{\mathrm{x}}^2\right)}^5}\mathrm{dx}\\ =\int \frac{{\sin }^7\mathrm{\theta }}{{\left(1-{\sin }^2\mathrm{\theta }\right)}^5}\cos \mathrm{\theta d\theta },\text{ where }\mathrm{x}=\sin \mathrm{\theta }\\ =\int {\tan }^7{\mathrm{\theta sec}}^2\mathrm{\theta d\theta }=\frac{{\tan }^8\mathrm{\theta }}{8}+\mathrm{C}=\frac{1}{8}\frac{{\mathrm{x}}^8}{{\left(1-{\mathrm{x}}^2\right)}^4}+\mathrm{C}\\ \mathrm{\lambda }=8\end{array}$