Answer :
The integral is ∫ x sec x tan x dx. We can solve it using integration by parts xln|sec(x) + tan(x)| - x + C
We have u = x, dv = sec(x)tan(x)dx, du = dx, and v = ln|sec(x) + tan(x)|.
The integral is then
Integrating the second part we get:
xln|sec(x) + tan(x)| - x + C
The integral is ∫ x sec x tan x dx. We can solve it using integration by parts. We let u = x and dv = sec(x)tan(x)dx. Then, du = dx and v = ln|sec(x) + tan(x)|. So, the integral becomes xln|sec(x) + tan(x)| - ∫ ln|sec(x) + tan(x)| dx. We can integrate the second part to get xln|sec(x) + tan(x)| - x + C.
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