A) \[\cos \,\,2x+c\]
B) \[2\,\cos \,\,2x+c\]
C) \[-\,\cos \,\,2x+c\]
D) \[-2\,\cos \,\,2x+c\]
Correct Answer: C
Solution :
\[f(x)=\underset{y\to x}{\mathop{\lim }}\,\,\,\,\frac{{{\sin }^{2}}\,y-\,{{\sin }^{2}}x}{{{y}^{2}}-{{x}^{2}}}\] \[\left[ \frac{0}{0}\,form \right]\] \[=\underset{y\to x}{\mathop{\lim }}\,\,\frac{2\sin \,y\,\cos \,y-0}{2y-0}\] \[=\frac{\sin \,2x}{2x}\] \[\therefore \]\[\int{4x\,\,f(x)\,\,dx=\int{4x\,\,\left( \frac{\sin 2x}{2x} \right)}\,\,dx}\] \[=2\int{\sin \,\,2x\,\,dx}\] \[=-\cos \,2x+c\]You need to login to perform this action.
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