BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    \[\int_{{}}^{{}}{{{\{1+2\tan x(\tan x+\sec x)\}}^{1/2}}dx}\]is equal to:

    A) \[log(sec\text{ }x+tan\text{ }x)+c\]                  

    B)                  \[log{{(sec\text{ }x+tan\text{ }x)}^{1/2}}+\text{ }c\]               

    C)                  \[~log\,sec\text{ }x\,(sec\text{ }x+tan\text{ }x)+c\]

    D)                   none of the above

    Correct Answer: C

    Solution :

    Let \[I={{\int_{{}}^{{}}{(1+2\tan x(\tan x+\sec x)\}}}^{1/2}}dx\] \[=\int_{{}}^{{}}{{{\{1+2{{\tan }^{2}}x+2\tan x\sec x\}}^{1/2}}}dx\] \[=\int_{{}}^{{}}{{{\{{{\sec }^{2}}x+{{\tan }^{2}}x+2\tan x\sec x\}}^{1/2}}dx}\] \[=\int_{{}}^{{}}{(\sec x+\tan x)dx}\] \[=\log (\sec x+\tan x)+\log \sec x+c\] \[=\log \sec x(\sec x+\tan x)+c\]


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