Question

Use integration by parts to integrate
x sec x tan x dx​

Answer 1

Answer:

Please see the attachment

Answer 2

$\displaystyle\longrightarrow\sf{\int x\ \sec x\ \tan x\ dx=\int x\ (\sec x\ \tan x)\ dx}$

By product rule of integration,

$\displaystyle\longrightarrow\sf{\int x\ \sec x\ \tan x\ dx=x\int(\sec x\ \tan x)\ dx-\int\dfrac{dx}{dx}\left[\int(\sec x\ \tan x)\ dx\right]\ dx}$

$\displaystyle\longrightarrow\sf{\int x\ \sec x\ \tan x\ dx=x\ \sec x-\int\sec x\ dx}$

$\displaystyle\longrightarrow\sf{\int x\ \sec x\ \tan x\ dx=x\ \sec x-\int\dfrac{\sec x(\sec x+\tan x)}{\sec x+\tan x}\ dx}$

$\displaystyle\longrightarrow\sf{\int x\ \sec x\ \tan x\ dx=x\ \sec x-\int\dfrac{\sec x\tan x+\sec^2x}{\sec x+\tan x}\ dx}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int x\ \sec x\ \tan x\ dx=x\ \sec x-\ln|\sec x+\tan x|+c}}}$