A) \[\frac{2}{\pi +1}\]
B) \[\frac{2}{\pi +2}\]
C) \[\frac{2}{\pi +4}\]
D) 0
Correct Answer: C
Solution :
\[N=\frac{1}{2}\,\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin \,2x}{{{(x+1)}^{2}}}dx}\] On putting \[2x=1,\] we get \[N=\frac{1}{4}.\,4\int\limits_{0}^{\frac{\pi }{2}}{\,\frac{\sin t}{{{(t+2)}^{2}}}dt}\] \[\left. N=(\sin t)\,\left( \frac{-1}{t+2} \right) \right]_{0}^{\frac{\pi }{2}}+\int\limits_{0}^{\frac{\pi }{2}}{\frac{\cos t}{t+2}dt}\] \[\left. \therefore \,\,M-N\,=\frac{\sin t}{\operatorname{t}+2} \right]_{0}^{\frac{\pi }{2}}\,=\frac{2}{\pi +4}\]
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