A) \[{{\log }_{e}}({{e}^{-x}}+\cos x)+C\]
B) \[{{\log }_{e}}(1+{{e}^{x}}\cos x)+C\]
C) \[{{\log }_{e}}({{e}^{-x}}+sinx)+C\]
D) \[{{\log }_{e}}(1+{{e}^{x}}sinx)+C\]
Correct Answer: B
Solution :
[b] \[\int{\frac{\cos 2x}{\left( {{e}^{-x}}+\cos x \right)\sqrt{1+\sin 2x}}}dx\] \[=\int{\frac{({{\cos }^{2}}x-{{\sin }^{2}}x)}{\left( {{e}^{-x}}+\cos x \right)\sqrt{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}}}\,dx\] \[=\int{\frac{\cos x-\sin x}{{{e}^{-x}}+\cos x}}dx\] \[=\int{\frac{{{e}^{x}}(\cos x-\sin x)}{1+{{e}^{x}}\cos x}}dx\] \[={{\log }_{e}}(1+{{e}^{x}}\cos x)+C\]
You need to login to perform this action.
You will be redirected in
3 sec
