A) \[\sin 2x+c\]
B) \[-\frac{1}{2}\sin 2x+c\]
C) \[\frac{1}{2}\sin 2x+c\]
D) \[-\sin 2x+c\]
Correct Answer: B
Solution :
Let\[\int{=\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}\] \[=\int{\frac{({{\sin }^{4}}x-{{\cos }^{4}}x)({{\sin }^{4}}x+{{\cos }^{4}}x)}{({{\sin }^{2}}x+{{\cos }^{2}}x)-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[=\int{\frac{({{\sin }^{4}}x-{{\cos }^{4}}x)({{\sin }^{4}}x+{{\cos }^{4}}x)}{({{\sin }^{4}}x+{{\cos }^{4}}x)}dx}\] \[=\int{({{\sin }^{4}}x-{{\cos }^{4}}x)dx}\] \[=\int{({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{2}}x+{{\cos }^{2}}x)dx}\] \[=-\int{({{\cos }^{2}}x-{{\sin }^{2}}x)}dx\] \[=-\int{\cos 2\,\,x\,\,dx}\] \[=-\frac{\sin 2x}{2}+c\]
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