A) \[{{\sin }^{n-1}}x\cos x\]
B) \[{{\cos }^{n-1}}x\sin x\]
C) \[-{{\sin }^{n-1}}x\cos x\]
D) \[-{{\cos }^{n-1}}x\sin x\]
Correct Answer: C
Solution :
We know that, if \[{{I}_{n}}=\], then \[{{I}_{n}}=-\frac{{{\sin }^{n-1}}x\cos x}{n}+\frac{n-1}{n}{{I}_{n-2}}\] where\[n\]is a positive integer. \[\Rightarrow \]\[n{{I}_{n}}-(n-1){{I}_{n-2}}=-{{\sin }^{n-1}}x\cos x\]
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