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Q12  Integrate the functions x \sec ^2 x

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Consider x \sec ^2 x

So, we have then: I =\int x\sec^2 x dx

After taking x as a first function and \sec^2x as second function and integrating by parts, we get

I =x\int \sec^2 x dx -\int \left \{ \left ( \frac{d}{dx}x \right )\int \sec^2 x dx \right \}dx

= x\tan x -\int1.\tan x dx

= x\tan x +\log|\cos x | +C

Posted by

Divya Prakash Singh

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