Integral Calculus Question 81
Question: $ \int \frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx $ is equal to
Options:
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 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \int{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx} $ $ =\int{\frac{(sin^{2}x-cos^{2}x)(sin^{4}x+cos^{4}x)}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}} $ $ =\int{-\cos 2xdx=-\frac{1}{2}\sin 2x+C} $
BETA
