A) \[2[\sin x+x\cos \alpha ]+c\]
B) \[2[\sin x+\sin \alpha ]+c\]
C) \[2[-\sin x+x\cos \alpha ]+c\]
D) \[-2[\sin x+\sin \alpha ]+c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }}\,dx=\int_{{}}^{{}}{\frac{2({{\cos }^{2}}x-{{\cos }^{2}}\alpha )}{\cos x-\cos \alpha }\,dx}\] \[=2\int_{{}}^{{}}{(\cos x+\cos \alpha )\,dx}=2(\sin x+x\cos \alpha )\].
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