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 :
\ t \[I=\int_{{}}^{{}}{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[=\int_{{}}^{{}}{\frac{{{({{\sin }^{4}}x)}^{2}}-{{({{\cos }^{4}}x)}^{2}}}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[=\int_{{}}^{{}}{\frac{({{\sin }^{4}}x+{{\cos }^{4}}x)(si{{n}^{4}}x-{{\cos }^{4}}x)}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[[{{(si{{n}^{2}}x+co{{s}^{2}}x)}^{2}}-2si{{n}^{2}}xco{{s}^{2}}x]\] \[=\int_{{}}^{{}}{\frac{[({{\sin }^{2}}x+{{\cos }^{2}}x][si{{n}^{2}}x-co{{s}^{2}}x]}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}dx\] \[=-\int_{{}}^{{}}{\cos 2xdx=\frac{-\sin 2x}{2}}+c=-\frac{1}{2}\sin 2x+c\]
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