JCECE Engineering JCECE Engineering Solved Paper-2010

  • question_answer
    \[\int{{{\sec }^{3}}x}\,\,dx\]is equal to

    A) \[\sec x\tan x+\log |\sec x+\tan x|+c\]

    B) \[\frac{1}{2}\sec x\tan x+\frac{1}{2}\log |\sec x+\tan x|+c\]

    C) \[{{\sec }^{3}}x\tan x+\log |{{\sec }^{3}}x+\tan x|+c\]

    D)  None of the above

    Correct Answer: B

    Solution :

    Let\[I=\int{{{\sec }^{2}}x}\,\,dx\] \[\Rightarrow \]               \[I=\int{\sec x}\cdot {{\sec }^{2}}x\,\,dx\] \[\Rightarrow \]               \[I=\sec x\tan x-\int{\sec x\tan x\cdot \tan x\,\,dx}\]                                                 (integrating by parts)                 \[I=\sec x\tan x-\int{\sec x}{{\tan }^{2}}x\,\,dx\]                    \[=\sec x\tan x-\int{\sec x({{\sec }^{2}}x-1)dx}\] \[\Rightarrow \]               \[I=\sec x\tan x-\int{({{\sec }^{2}}x-\sec x)dx}\] \[\Rightarrow \]               \[I=\sec x\tan x-\int{{{\sec }^{2}}x\,\,dx+\int{\sec x\,\,dx}}\] \[\Rightarrow \]               \[2I=\sec x\tan x+\int{\sec x\,\,dx}\] \[\Rightarrow \]               \[2I=\sec x\tan x+\log |\sec x+\tan x|+c\] \[\Rightarrow \]    \[I=\frac{1}{2}\sec x\tan x+\frac{1}{2}\log |\sec x+\tan x|+c\]


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