A) \[0\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) \[2\pi \]
Correct Answer: D
Solution :
| Let \[I=\int_{-\pi }^{\pi }{{{(\cos \alpha x-\sin \beta x)}^{2}}dx}\] |
| \[\Rightarrow \]\[I=\int_{-\pi }^{\pi }{({{\cos }^{2}}\alpha x+{{\sin }^{2}}\beta -2\sin \beta x\cos \alpha x)dx}\] |
| \[\Rightarrow \]\[I=2\int_{0}^{\pi }{({{\cos }^{2}}\alpha x+{{\sin }^{2}}\beta x)dx}\] |
| \[[\because \sin \beta x\cos \alpha x\,\text{is}\,\text{an}\,\text{odd function }\!\!]\!\!\text{ }\] |
| \[\Rightarrow \]\[I=2\int_{0}^{\pi }{\left[ \frac{1+\cos 2\alpha x}{2}+\frac{1-\cos 2\beta x}{2} \right]}dx\] |
| \[\Rightarrow \]\[I=2\left[ x+\frac{\sin 2\alpha x}{2x}-\frac{\sin 2\beta x}{2\beta } \right]_{0}^{\pi }\] |
| \[\Rightarrow \]\[I=2\,[\pi ]=2\pi \] |
You need to login to perform this action.
You will be redirected in
3 sec
