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 \[{{I}_{n}}=\int{{{\sin }^{n}}x\,dx,\,\text{then}}\] \[{{I}_{n}}=-\frac{{{\sin }^{n-1}}x\,\cos x}{n}+\frac{n-1}{n}{{I}_{n-2}}\] where n is a positive integer. \[\Rightarrow \]\[n{{I}_{n}}-(n-1){{I}_{n-2}}=-{{\sin }^{n-1}}x\cos x\]
You need to login to perform this action.
You will be redirected in
3 sec
