A) \[\frac{1}{2}\sin 2x+C\]
B) \[-\frac{1}{2}\sin 2x+C\]
C) \[-\frac{1}{2}\sin x+C\]
D) \[-{{\sin }^{2}}x+C\]
Correct Answer: B
Solution :
\[I=\int{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[=\int{\frac{({{\sin }^{4}}x-{{\cos }^{4}}x)\,({{\sin }^{4}}x+{{\cos }^{4}}x)}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[=\int{\frac{\begin{align} & ({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{2}}x+{{\cos }^{2}}x) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{\sin }^{4}}x+{{\cos }^{4}}x) \\ \end{align}}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[=\int{\frac{\begin{align} & 1.({{\sin }^{2}}x-{{\cos }^{2}}x)[{{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2{{\sin }^{2}}x{{\cos }^{2}}x] \\ \end{align}}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}\] \[=\int{\frac{({{\sin }^{2}}x-{{\cos }^{2}}x)(1-2si{{n}^{2}}{{\cos }^{2}}x)}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[=-\int{\cos \,2x\,\,dx=-\frac{1}{2}}\sin \,2x+C\]
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