A) 2007
B) 2006
C) 2009
D) 2008
Correct Answer: D
Solution :
\[{{\theta }_{1}}+{{\theta }_{2}}=\frac{\pi }{2}\] \[\therefore \]\[I=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{\frac{d\theta }{1+\tan \left( \frac{\pi }{2}-\theta \right)}}=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{\frac{\tan \theta d\theta }{1+\tan \theta }}\] And also\[I=\int_{\theta 1}^{\theta 2}{\frac{d\theta }{1+\sin \theta }}\] \[\therefore \]\[2I=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{d\theta ={{\theta }_{2}}-{{\theta }_{1}}=\frac{1002\pi }{2008}\Rightarrow I=\frac{50l\pi }{2008}}\] Hence, K= 2008.
You need to login to perform this action.
You will be redirected in
3 sec
