Integral Calculus Question 265
Question: $ \int_{{}}^{{}}{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}\ dx=} $
[IIT 1986]
Options:
A) $ \sin 2x+c $
B) $ -\frac{1}{2}\sin 2x+c $
C) $ \frac{1}{2}\sin 2x+c $
D) $ -\sin 2x+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \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}}-2{{\sin }^{2}}x{{\cos }^{2}}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_{{}}^{{}}{({{\sin }^{2}}x-{{\cos }^{2}}x),dx} $ $ =\int_{{}}^{{}}{-\cos 2x,dx=-\frac{\sin 2x}{2}+c.} $
BETA
