A) \[\frac{K}{2}\]
B) \[\frac{\pi }{2}-K\]
C) \[\frac{\pi }{2}+K\]
D) \[\frac{\pi }{2}-2K\]
Correct Answer: D
Solution :
| We have, \[K=\int_{0}^{\pi /2}{\frac{\sin x}{1+\sin x+\cos x}dx}\] |
| \[\Rightarrow \]\[K=\int_{0}^{\pi /2}{\frac{\sin \left( \frac{\pi }{2}-x \right)}{1+\sin \left( \frac{\pi }{2}-x \right)+\cos \left( \frac{\pi }{2}-x \right)}dx}\] |
| \[\Rightarrow \]\[K=\int_{0}^{\pi /2}{\frac{\cos x}{1+\cos x+\sin x}dx}\] |
| \[\Rightarrow \]\[2K=\int_{0}^{\pi /2}{\frac{\sin x+cox}{1+\cos x+\sin x}dx}\] |
| \[\Rightarrow \]\[2K=\int_{0}^{\pi /2}{\frac{1+\sin x+\cos x}{1+\cos x+\sin x}dx}\]\[-\int_{0}^{\pi /2}{\frac{dx}{1+\sin x+\cos x}}\] |
| \[\Rightarrow \]\[\int_{0}^{\pi /2}{\frac{dx}{1+\sin x+\cos x}=\frac{\pi }{2}-2K}\] |
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