Integral Calculus Question 212
Question: $ \int{( {{\sin }^{4}}x-{{\cos }^{4}}x ),dx=} $
[RPET 2003]
Options:
A) $ -\frac{\cos 2x}{2}+c $
B) $ -\frac{\sin 2x}{2}+c $
C) $ \frac{\sin 2x}{2}+c $
D) $ \frac{\cos 2x}{2}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \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 }^{2}}x-{{\sin }^{2}}x)dx} $ $ =-\int_{{}}^{{}}{\cos 2x,dx} $ $ =\frac{-\sin 2x}{2}+c $ .
BETA
