A) \[{{2}^{-n+1}}\]
B) \[{{2}^{n+1}}\]
C) \[{{2}^{-n\,-1}}\]
D) \[{{2}^{-\,n}}\]
Correct Answer: D
Solution :
[d]| We have, \[I=\int_{0}^{\pi /2}{{{\cos }^{n}}x{{\sin }^{n}}dx}\] |
| \[I=\int_{0}^{\pi /2}{\frac{{{(2\sin x\cos x)}^{n}}}{{{2}^{n}}}dx}\]\[\Rightarrow \]\[I=\frac{1}{{{2}^{n}}}\int_{0}^{\pi /2}{{{(\sin 2x)}^{n}}dx}\] \[\left[ \text{put}\,2x=t\Rightarrow dx=\frac{dt}{2} \right]\] |
| \[\Rightarrow \]\[I=\frac{1}{{{2}^{n}}}\int_{0}^{\pi }{{{(\sin t)}^{n}}\frac{dt}{2}}\] |
| \[\Rightarrow \]\[I=\frac{1}{{{2}^{n+1}}}\int_{0}^{\pi }{{{(\sin t)}^{n}}dt}\] |
| \[\Rightarrow \]\[I=\frac{2}{{{2}^{n+1}}}\int_{0}^{\pi /2}{{{\sin }^{n}}tdt}\] \[\left[ \because \int_{0}^{2a}{f(x)dx=2\int_{0}^{a}{f\,(2a-x)dx}} \right]\] |
| \[\Rightarrow \]\[I={{2}^{-n}}\int_{0}^{\pi /2}{{{\sin }^{n}}x\,dx}\] |
| \[\therefore \]\[\lambda ={{2}^{-n}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
