What will be the value of this? Integration secxd(sec x)

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Answer 1 · 2018-02-09T13:46:26

${sec}^{2} x - \frac{1}{2} {tan}^{2} x + C$ $sec^2 x - 1/2 tan^2 x +C$

This is integration of the product of two functions. One is sec x and the other is $\frac{d}{d x} (sec x)$ $d/dx (secx)$

Let first function be secx and the second function be $\frac{d}{d x} (sec x)$ $d/dx (sec x)$ .

Application of the product rule of integration is:

first function * integral of second - $\int$ $int$ differentiation of first *integral of second

Now applying this rule would result in following:

sec x * sec x - $\int$ $int$ secx tanx * sec x dx

= ${sec}^{2} x - \int {sec}^{2} x tan x d x$ $sec^2 x - int sec^2 x tan x dx$ Hint: [let tan x =t so that sec^ x dx =dt]

= ${sec}^{2} x - \frac{1}{2} {tan}^{2} x + C$ $sec^2 x - 1/2 tan^2 x +C$