A) \[{{e}^{2x}}\tan x+c\]
B) \[{{e}^{2x}}\cot x+c\]
C) \[\frac{{{e}^{2x}}\tan x}{2}+c\]
D) \[\frac{{{e}^{2x}}\cot x}{2}+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{{{e}^{2x}}\frac{1+\sin 2x}{1+\cos 2x}\,dx}=\int_{{}}^{{}}{{{e}^{2x}}\left[ \frac{1}{1+\cos 2x}+\frac{\sin 2x}{1+\cos 2x} \right]\,dx}\] \[=\int_{{}}^{{}}{{{e}^{2x}}\left[ \frac{{{\sec }^{2}}x}{2}+\tan x \right]}\,dx\] \[=\frac{1}{2}\int_{{}}^{{}}{{{e}^{2x}}{{\sec }^{2}}x\,dx}+\int_{{}}^{{}}{{{e}^{2x}}\tan x\,dx}\] \[=\frac{{{e}^{2x}}\tan x}{2}-\int_{{}}^{{}}{\frac{{{e}^{2x}}{{\sec }^{2}}x}{2}\,dx}+\int_{{}}^{{}}{\frac{{{e}^{2x}}{{\sec }^{2}}x}{2}\,dx}+c\] \[=\frac{{{e}^{2x}}\tan x}{2}+c.\]
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