A) \[\tan ({{e}^{x}})-x+c\]
B) \[{{e}^{x}}(\tan {{e}^{x}}-1)+c\]
C) \[\sec ({{e}^{x}})+c\]
D) \[\tan ({{e}^{x}})-{{e}^{x}}+c\]
Correct Answer: D
Solution :
Put \[{{e}^{x}}=t\Rightarrow {{e}^{x}}dx=dt,\] then \[\int_{{}}^{{}}{{{e}^{x}}{{\tan }^{2}}({{e}^{x}})\,dx}=\int_{{}}^{{}}{{{\tan }^{2}}t\,dt}=\int_{{}}^{{}}{({{\sec }^{2}}t-1)\,dt}\] \[=\tan t-t+c=\tan ({{e}^{x}})-{{e}^{x}}+c.\]
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