A) \[-\frac{\cos 2x}{2}+c\]
B) \[-\frac{\sin 2x}{2}+c\]
C) \[\frac{\sin 2x}{2}+c\]
D) \[\frac{\cos 2x}{2}+c\]
Correct Answer: B
Solution :
\[\int{({{\sin }^{4}}x-{{\cos }^{4}}x)dx}=\int{({{\sin }^{2}}x-{{\cos }^{2}}x)}\,({{\sin }^{2}}x+{{\cos }^{2}}x)\,dx\] \[=\int{({{\sin }^{2}}x-{{\cos }^{2}}x)\,dx}\]\[=-\int_{{}}^{{}}{({{\cos }^{2}}x-{{\sin }^{2}}x)dx}\] \[=-\int_{{}}^{{}}{\cos 2x\,dx}\]\[=\frac{-\sin 2x}{2}+c\].
You need to login to perform this action.
You will be redirected in
3 sec
